Re: regular expression, this but not that
On 9/18/2011 9:38 PM, Jer A wrote: All, lets say i want to test if a string contains pig but not dog and/or not cat i tried something like this: =~ m/[^dog]|[^cat]|pig/ig what is the best way of going about this, using one regex? your help is very much appreciated, perhaps the following: if ($s =~ m/((dog|cat)?.*pig.*(dog|cat)?)/ig $1 == pig) ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: regular expression, this but not that
On Mon, Sep 19, 2011 at 7:38 AM, Jer A jeremy...@hotmail.com wrote:
All,
lets say i want to test if a string contains pig but not dog and/or not
cat
i tried something like this:
=~ m/[^dog]|[^cat]|pig/ig
what is the best way of going about this, using one regex?
your help is very much appreciated,
thanks.
Hi Jeremy,
I am not sure why would you want to use only one regex
and IMHO it would be very complex to write that.
It would make your code both unreadable and slow.
I'd use two regexes for that:
if ($str =~ m/pig/ and $str !~ /dog|cat/) {
}
Add the railing /i if you want this to be case insensitive.
The /g is not needed as the first pig will be ok
and the first dog or cat would also be indication that the match should fail.
regards
Gabor
--
Gabor Szabo http://szabgab.com/
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RE: Regular expression with variable
I have a regular expression problem how do i use scalar variables in substitution and complex matching? eg I want the following to work. $string =~ s/^$variable//; $string =~ m/^([^$variable]*)/; thanks in advance for your help. -Jeremy A. $x = Cowboy; $y = ow; $x =~ s/$y/-/; print $x; ; # Output is C-boy _ Get FREE company branded e-mail accounts and business Web site from Microsoft Office Live http://clk.atdmt.com/MRT/go/mcrssaub0050001411mrt/direct/01/ ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular expression question
Title: Regular expression question From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Cai, Lucy (L.)Sent: Monday, July 31, 2006 17:21To: Cai, Lucy (L.); perl-win32-users@listserv.ActiveState.com; perl-unix-users@listserv.ActiveState.com; [EMAIL PROTECTED]; [EMAIL PROTECTED]Subject: Regular _expression_ question I have a file such as: My $file = "c:\temp\zips\ok.txt"; How can I split the $file to get the only path: My $dir = "c:\temp\zips"; My $file = "ok.txt"; Thanks in advance! Lucy Better to use File::Basename which will break it out foryou. Comes as part of std Perl. Wags ;) ** This message contains information that is confidential and proprietary to FedEx Freight or its affiliates. It is intended only for the recipient named and for the express purpose(s) described therein. Any other use is prohibited. ** ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular expression question
Title: Regular expression question Cai, Lucy (L.) wrote, on Monday, July 31, 2006 8:21 PM : My $file = "c:\temp\zips\ok.txt"; : How can I split the $file to get the only path: : My $dir = "c:\temp\zips";: My $file = "ok.txt"; May I suggest you use File:Basename instead of a regex? Joe Joseph Discenza, Senior Programmer/Analystmailto:[EMAIL PROTECTED]Carleton Inc. http://www.carletoninc.com574.243.6040 ext. 300Fax: 574.243.6060 From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of : Cai, Lucy (L.); perl-win32-users@listserv.ActiveState.com; perl-unix-users@listserv.ActiveState.com; [EMAIL PROTECTED]; [EMAIL PROTECTED]Subject: Regular _expression_ question I have a file such as: Thanks in advance! Lucy ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular expression question
At 09:45 PM 4/26/2006 -0400, Cai, Lucy \(L.\) wrote: return (($Output =~ /.*\(ucmvob\)/s*$/) ? 1 : 0); $Output =/vobs/na_mscs_pvob /ccstore/ecc/vobs_fcis321/na_mscs_pvob.vbs public (ucmvob,replicated) What I want to do is if this tring include word ucmvob, then return 1, else return 0. If u just want the word ucmvob then u don't need any of the rest of the regex u put in there. And I think the outer parenthesis on the return could be messing u up too. Not sure without testing but that could be trying to return an array. This will work: $Output =~ m/ucmvob/ ? return 1 : return 0; If u want to retain ur previous return format then do this: return scalar $Output =~ m/ucmvob/; # returns no. of matches i.e. 1 or 0 -- REMEMBER THE WORLD TRADE CENTER ---= WTC 911 =-- ...ne cede malis 0100 ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular expression question
Well, there are a couple of issues here. First off, I don't think this
would even compile, because you used /s instead of \s. Secondly, your
regex is looking for:
.* zero or more of any character
(unnecessary, since you didn't anchor the start of the
string)
\(ucmvob\) the string '(ucmvob)'
\s* zero or more whitespace characters
$ the end of the string
--
I think what you might be looking for is something similar to this:
/\(ucmvob\)/
or possibly this:
/\(ucmvob,.*?\)/
which would match 'ucmvob', a comma, and any other text (non-greedy
match) between parentheses.
Of course, I can't test this where I am, but that should get you
started.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Cai, Lucy (L.)
Sent: Wednesday, April 26, 2006 6:46 PM
To: perl-win32-users@listserv.ActiveState.com
Cc: perl-win32-users@listserv.ActiveState.com
Subject: Regular expression question
Hi, all,
I have a question about regular expression:
my code is like this:
**
sub IsPVob
{
my ($Vob) = @_;
my $Output = `cleartool lsvob $Vob`;
die IsPVob can't list vob $Vob if $?;
return (($Output =~ /.*\(ucmvob\)/s*$/) ? 1 : 0);
}
**
$Output =/vobs/na_mscs_pvob
/ccstore/ecc/vobs_fcis321/na_mscs_pvob.vbs public (ucmvob,replicated)
What I want to do is if this tring include word ucmvob, then return 1,
else return 0.
My code does not work, it return 0. Could you please see why it is like
this? or do you have a better idea?
Thanks a lot inadvance!
Lucy
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Re: Regular expression question
Cai, Lucy (L.) [EMAIL PROTECTED] graced perl with these words of wisdom: return (($Output =~ /.*\(ucmvob\)/s*$/) ? 1 : 0); } ** $Output =/vobs/na_mscs_pvob /ccstore/ecc/vobs_fcis321/na_mscs_pvob.vbs public (ucmvob,replicated) What I want to do is if this tring include word ucmvob, then return 1, else return 0. If I understand your code correctly, you're looking for the string (ucmvob) -- note that you don't have anything in your regex between the b and the escaped close partentheses. In $Output, that string doesn't exist, as the b in ucmvob is followed by a comma. -- Ted fedya at bestweb dot net Oh Marge, anyone can miss Canada, all tucked away down there --Homer Simpson ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular Expression
$String = 'Integration Test Lead: \ul\b0 \tab\tab\tab Kimberly Kim\tab\tab\tab\tab\tab\tab \par'; How would I extract Kimberly Kim from the string using regular expression. Is there a way using expression to skip \[characters] ( like \tab ) and search for only real words, maybe between spaces. Have you thought of using the \t\t\t as a marker for the start of the string you want to capture, like: $string =~ /\t\t\t([^\t]+)/ ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular Expression
At 05:40 PM 11/28/2005 -0800, Wong, Danny H. wrote: $String = 'Integration Test Lead: \ul\b0 \tab\tab\tab Kimberly Kim\tab\tab\tab\tab\tab\tab \par'; How would I extract Kimberly Kim from the string using regular expression. Is there a way using expression to skip \[characters] ( like \tab ) and search for only real words, maybe between spaces. So u want to be able to search for multi word strings, that may or may not be interrupted by whitespace or other junk characters? $match = $string =~ m/(Kimberly\W*\s\W*Kim)/s; @match = $string =~ m/(Kimberly)\W*\s\W*(Kim)/s; @match = $string =~ m/($firstname)\W*\s\W*($lastname)/s; ($firstname, $lastname) = $string =~ m/($firstname)\W*\s\W*($lastname)/s; \s is whitespace and \W is any non alphanumeric garbage. If any high ASCII characters are valid name letters it's a little more complicated but basically the same format. -- REMEMBER THE WORLD TRADE CENTER ---= WTC 911 =-- ...ne cede malis 0100 ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular expression
Don't know if this works, but have you tried: $string = 1\\.2\\.3; - Original Message - From: Wong, Danny H. [EMAIL PROTECTED] To: Sisyphus [EMAIL PROTECTED]; Jan Dubois [EMAIL PROTECTED]; perl-win32-users perl-win32-users@listserv.ActiveState.com Sent: Thursday, September 15, 2005 2:28 PM Subject: Regular expression Hi Perl Gurus, I have a regular expression question. I have a variable $Number = 1.2.3.4 When I use the variable $Number as part of my regular expression, the . character gets interpret as any character. How do I make it a literal . that I'm searching for? Example: $String = This is a numeric string 1.2.3.411 embedded within another string 3.4.5.6 $String =~ m/$Number/I; This returns 1.2.3.411 true, when it shouldn't. Thanks for your help! Thanks Danny Wong SCM Engineer PowerTV Inc., - - - - - - - Appended by PowerTV, A division of Scientific Atlanta. - - - - - - - This e-mail and any attachments may contain information that is confidential, proprietary, privileged or otherwise protected by law. The information is solely intended for the named addressee (or a person responsible for delivering it to the addressee). If you are not the intended recipient of this message, you are not authorized to read, print, retain, copy or disseminate this message or any part of it. If you have received this e-mail in error, please notify the sender immediately by return e-mail and delete it from your computer. ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular expression
Hello Danny, Thursday, September 15, 2005, 9:28:44 AM, You wrote: WDH $Number = 1.2.3.4 WDH When I use the variable $Number as part of my regular expression, the WDH . character gets interpret as any character. How do I make it a WDH literal . that I'm searching for? WDH Example: WDH $String = This is a numeric string 1.2.3.411 embedded within another WDH string 3.4.5.6 WDH $String =~ m/$Number/I; This returns 1.2.3.411 true, when it WDH shouldn't. If You mean that '1.2.3.4' is a word i.e. has whitespaces before and after it, than your pattern have to be $String =~ /$Number\b/; That match ' 1.2.3.4 ' and do not match '1.2.3.411'. Dots in $Number do not affect matching because variable substituted as string. And note: 'i' modifier MUST be lowercase. -- Best regards, Сергейmailto:[EMAIL PROTECTED] ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular expression
Wong, Danny H. wrote: Hi Perl Gurus, I have a regular expression question. I have a variable $Number = 1.2.3.4 When I use the variable $Number as part of my regular expression, the . character gets interpret as any character. How do I make it a literal . that I'm searching for? Example: $String = This is a numeric string 1.2.3.411 embedded within another string 3.4.5.6 $String =~ m/$Number/I; This returns 1.2.3.411 true, when it shouldn't. Easiest way is $String =~ m/\Q$Number\E/i; You can also use quotemeta (look it up) or escape your .'s in $number (ie: 1\.2\.3\.4). PS: It's not good etiquette to CC people who already read the list. -- ,-/- __ _ _ $Bill LuebkertMailto:[EMAIL PROTECTED] (_/ / )// // DBE CollectiblesMailto:[EMAIL PROTECTED] / ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/ -/-' /___/__/_/_http://dbecoll.tripod.com/ (My Perl/Lakers stuff) ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular expression
$Bill Luebkert wrote: You can also use quotemeta (look it up) or escape your .'s in $number (ie: 1\.2\.3\.4). That should have been '1\.2\.3\.4' or 1\\.2\\.3\\.4. -- ,-/- __ _ _ $Bill LuebkertMailto:[EMAIL PROTECTED] (_/ / )// // DBE CollectiblesMailto:[EMAIL PROTECTED] / ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/ -/-' /___/__/_/_http://dbecoll.tripod.com/ (My Perl/Lakers stuff) ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular Expression Help Please
-Original Message-
One of the columns I'm calling out of my database is the email one. I'ts
in MAPI format,
so I need to extract the very last part. So the bottom example I would
need to grab
JDoe
MAPI:{Doe, John}EX:/o=Company/ou=Site/cn=Recipients/cn=JDoe
Steve,
An easy way might be:
$_ = 'MAPI:{Doe, John}EX:/o=Company/ou=Site/cn=Recipients/cn=JDoe';
/([^=]+)$/;
print $1;
Chris
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RE: Regular Expression Help Please
From: [EMAIL PROTECTED] wrote:
: One of the columns I'm calling out of my database is the email
: one. I'ts in MAPI format, so I need to extract the very last
: part. So the bottom example I would need to grab
:
: JDoe
:
: MAPI:{Doe, John}EX:/o=Company/ou=Site/cn=Recipients/cn=JDoe
:
:
: Then I can append the rest as
:
: [EMAIL PROTECTED]
:
: Every one is the same format,
:
: MAPI:{Smith, Jane}EX:/o=Company/ou=Site/cn=Recipients/cn=LJSmith
:
: Any help greatly appreciated.
my $field = 'MAPI:{Smith,
Jane}EX:/o=Company/ou=Site/cn=Recipients/cn=LJSmith';
my $email;
if ( $field =~ /=([^=]+)$/ ) {
$email = [EMAIL PROTECTED];
} else {
# Uh Oh!
}
The '$' is a zero length placeholder for the end of
the string. [^=] is a character class of all characters
which are not the equal sign. Anything inside parenthesis
'()' will be captured to $1.
HTH,
Charles K. Clarkson
--
Mobile Homes Specialist
254 968-8328
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Re: Regular Expression Help Please
You can get that plus some other info with this regex:
$string = 'MAPI:{Doe, John}EX:/o=Company/ou=Site/cn=Recipients/cn=JDoe';
($lastname, $firstname, $mailid) = $string =~ m/^.+?\{(\w+),
(\w+)\}.+?cn\=(\w+)$/;
At 09:46 PM 2/8/05 +, steve silvers wrote:
One of the columns I'm calling out of my database is the email one. I'ts in
MAPI format, so I need to extract the very last part. So the bottom example
I would need to grab
JDoe
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RE: Regular Expression Help Please
-Original Message-
One of the columns I'm calling out of my database is the email one. I'ts in
MAPI format, so I need to extract the very last part. So the bottom example
I would need to grab
JDoe
MAPI:{Doe, John}EX:/o=Company/ou=Site/cn=Recipients/cn=JDoe
Then I can append the rest as
[EMAIL PROTECTED]
Every one is the same format,
MAPI:{Smith, Jane}EX:/o=Company/ou=Site/cn=Recipients/cn=LJSmith
Any help greatly appreciated.
Thanks in advance
Steve
---
I know there's probably a better way to do this, but...
my $field='MAPI:{Doe, John}EX:/o=Company/ou=Site/cn=Recipients/cn=JDoe';
my @bla = split(/=/, $field);
print pop(@bla);
- Chris
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RE: Regular expression to test for numeric values
So I get: /^-?(?:\d+\.?\d*|\.\d+)$/ I'm being thrown by the ?: What's that all about? R. ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: regular expression question
Thanks for the replying. I have another question, if I have a string like $a = this is a (test); How can I change it to $a = this_is_a_test; How can I remove ()? Thanks Lixin -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of $Bill Luebkert Sent: Wednesday, March 31, 2004 8:27 PM To: [EMAIL PROTECTED] Subject: Re: regular expression question Wagner, David --- Senior Programmer Analyst --- WGO wrote: Hey guys - what's with the HTML ? From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Wednesday, March 31, 2004 16:49 To: [EMAIL PROTECTED] Subject: regular expression question I have a $a which $a = this is a test; How can I change $a like: $a = this_is_a_test; No problem using $a, but if I were you, I would not. In the Perl sort $a, $b are the default variables to use and I would stay away if I were you. That said, $a =~ s/ /_/g; You could also use tr (slightly faster) : $a =~ tr/ /_/; -- ,-/- __ _ _ $Bill LuebkertMailto:[EMAIL PROTECTED] (_/ / )// // DBE CollectiblesMailto:[EMAIL PROTECTED] / ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/ -/-' /___/__/_/_http://dbecoll.tripod.com/ (My Perl/Lakers stuff) ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: regular expression question
$a = this is a (test); $a =~ s/\W+/_/g; HTH -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Thursday, April 01, 2004 11:28 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: regular expression question Thanks for the replying. I have another question, if I have a string like $a = this is a (test); How can I change it to $a = this_is_a_test; How can I remove ()? Thanks Lixin -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of $Bill Luebkert Sent: Wednesday, March 31, 2004 8:27 PM To: [EMAIL PROTECTED] Subject: Re: regular expression question Wagner, David --- Senior Programmer Analyst --- WGO wrote: Hey guys - what's with the HTML ? From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Wednesday, March 31, 2004 16:49 To: [EMAIL PROTECTED] Subject: regular expression question I have a $a which $a = this is a test; How can I change $a like: $a = this_is_a_test; No problem using $a, but if I were you, I would not. In the Perl sort $a, $b are the default variables to use and I would stay away if I were you. That said, $a =~ s/ /_/g; You could also use tr (slightly faster) : $a =~ tr/ /_/; -- ,-/- __ _ _ $Bill LuebkertMailto:[EMAIL PROTECTED] (_/ / )// // DBE CollectiblesMailto:[EMAIL PROTECTED] / ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/ -/-' /___/__/_/_http://dbecoll.tripod.com/ (My Perl/Lakers stuff) ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: regular expression question
$a =~ s/ /_/g; $a =~ s/[()]//g; or $a =~ tr/ ()/_/d; Peter Guzis Web Administrator, Sr. ENCAD, Inc. - A Kodak Company email: [EMAIL PROTECTED] www.encad.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Thursday, April 01, 2004 9:28 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: regular expression question Thanks for the replying. I have another question, if I have a string like $a = this is a (test); How can I change it to $a = this_is_a_test; How can I remove ()? Thanks Lixin -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of $Bill Luebkert Sent: Wednesday, March 31, 2004 8:27 PM To: [EMAIL PROTECTED] Subject: Re: regular expression question Wagner, David --- Senior Programmer Analyst --- WGO wrote: Hey guys - what's with the HTML ? From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Wednesday, March 31, 2004 16:49 To: [EMAIL PROTECTED] Subject: regular expression question I have a $a which $a = this is a test; How can I change $a like: $a = this_is_a_test; No problem using $a, but if I were you, I would not. In the Perl sort $a, $b are the default variables to use and I would stay away if I were you. That said, $a =~ s/ /_/g; You could also use tr (slightly faster) : $a =~ tr/ /_/; -- ,-/- __ _ _ $Bill LuebkertMailto:[EMAIL PROTECTED] (_/ / )// // DBE CollectiblesMailto:[EMAIL PROTECTED] / ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/ -/-' /___/__/_/_http://dbecoll.tripod.com/ (My Perl/Lakers stuff) ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: regular expression question
Stacy Doss wrote: $a = "this is a (test)"; $a =~ s/\W+/_/g; HTH -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of [EMAIL PROTECTED] Sent: Thursday, April 01, 2004 11:28 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: regular _expression_ question Thanks for the replying. I have another question, if I have a string like $a = "this is a (test)"; How can I change it to $a = "this_is_a_test"; How can I remove ()? Thanks Lixin -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]] On Behalf Of $Bill Luebkert Sent: Wednesday, March 31, 2004 8:27 PM To: [EMAIL PROTECTED] Subject: Re: regular _expression_ question Wagner, David --- Senior Programmer Analyst --- WGO wrote: Hey guys - what's with the HTML ? From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of [EMAIL PROTECTED] Sent: Wednesday, March 31, 2004 16:49 To: [EMAIL PROTECTED] Subject: regular _expression_ question I have a $a which $a = "this is a test"; How can I change $a like: $a = "this_is_a_test"; No problem using $a, but if I were you, I would not. In the Perl sort $a, $b are the default variables to use and I would stay away if I were you. That said, $a =~ s/ /_/g; You could also use tr (slightly faster) : $a =~ tr/ /_/; -- hi, you might want to add the following statement after words: $a =~ s/__/_/g; # for viewing reasons... regards, topdog "i have slipped the surly bonds of earth, and danced the skies on laughter-silvered wings;" --john gillespie magee jr. ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular expression to test for numeric values
$txtype = 0 unless $txtype =~ /^-{0,1}\d+(?:\.\d+){0,1}$/;
Peter Guzis
Web Administrator, Sr.
ENCAD, Inc.
- A Kodak Company
email: [EMAIL PROTECTED]
www.encad.com
-Original Message-
From: Motter, Jeffrey D [mailto:[EMAIL PROTECTED]
Sent: Thursday, April 01, 2004 10:40 AM
To: Perl-Win32-Users
Subject: Regular expression to test for numeric values
I've read the FAQ's on this, but they don't seem to answer the question.
I have a variable that could contain any value( alpha, alpha-numeric, or
numeric). If the value is NOT numeric, I need to change the variables' value
to 0( as in zero ).
Examples:
$txtype=2.314; # is numeric, so keep the value
$txtype=-2.314; # is numeric, so keep the value
$txtype=7; # is numeric, so keep the value
$txtype=-7; # is numeric, so keep the value
$txtype=2.31.4; # is not numeric, so change the value to 0
$txtype=7-7; # is not numeric, so change the value to 0
$txtype=UNKNOWN; # is not numeric, so change the value to 0
$txtype=7+E09; # is not numeric( even though it really is ), so change
the value to 0
My guess is that I need a regex that will match on any character that is:
not 0-9
or
more than one .
or
more than one -
or if - is not the first character of the string
Any ideas? Is it possible to do without using additional modules?
Thanks!
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RE: Regular expression to test for numeric values
-Original Message-
My guess is that I need a regex that will match on any character that is:
not 0-9
or
more than one .
or
more than one -
or if - is not the first character of the string
Any ideas? Is it possible to do without using additional modules?
--
I haven't written a regex in a few months, but how about something like
this:
for (2.314,-2.314,7,-7,2.31.4,7-7,UNKNOWN,7+E09) {
$txtype=$_;
$txtype=0 unless /^-?((\d+)|(\d*\.\d+)|(\d+\.\d*))$/;
print $_ = $txtype\n;
}
^ match beginning of string
-? string MIGHT begin with a -
(\d+) string of numbers with no .
(\d*\.\d+) string of numbers with a . (digits only required AFTER .)
(\d+\.\d*) string of numbers with a . (digits only required BEFORE .)
(..|..|..) pick one of the three options for a match
$ match end of the string
Chris
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RE: Regular expression to test for numeric values
Gerber, Christopher J wrote, on Thursday, April 01, 2004 2:12 PM
: -Original Message-
: My guess is that I need a regex that will match on any character that is:
: not 0-9
: or
: more than one .
: or
: more than one -
: or if - is not the first character of the string
:
: Any ideas? Is it possible to do without using additional modules?
: --
: I haven't written a regex in a few months, but how about something like
: this:
:
:for (2.314,-2.314,7,-7,2.31.4,7-7,UNKNOWN,7+E09) {
: $txtype=$_;
: $txtype=0 unless /^-?((\d+)|(\d*\.\d+)|(\d+\.\d*))$/;
: print $_ = $txtype\n;
:}
:
: ^match beginning of string
: -? string MIGHT begin with a -
: (\d+)string of numbers with no .
: (\d*\.\d+) string of numbers with a . (digits only required AFTER .)
: (\d+\.\d*) string of numbers with a . (digits only required BEFORE .)
: (..|..|..) pick one of the three options for a match
: $match end of the string
I'd make just a few suggestions. First, since there's no need for capture,
dispense with the internal capturing parentheses, and make the outer ones
non-capturing. This will speed up the regex engine a little, and if there's
a lot of stuff to go through...
Secondly, if you make your decimal optional in the third alternation,
it's no different from the first, so you can eliminate one of the alternatives.
Then you've also matched those with *any* digits before the decimal, so you
can remove the leading \d* from the other alternative.
So I get:
/^-?(?:\d+\.?\d*|\.\d+)$/
for the regex. Just a slight improvement (and a shorter regex is almost
always more readable, too). You could embed the comments in the regex
(use the /x flag) to make it crystal clear.
Joe
==
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RE: Regular expression to test for numeric values
Title: RE: Regular expression to test for numeric values
use Scalar::Util;
if (Scalar::Util::looks_like_number $num)
{
print Yes, $num is a number\n;
}
else
{
print no.\n;
}
... found it with perldoc -q number, which also showed a number of regex's that do the same kinda thing
-Original Message-
From: Motter, Jeffrey D [mailto:[EMAIL PROTECTED]]
Sent: Thursday, April 01, 2004 1:40 PM
To: Perl-Win32-Users
Subject: Regular _expression_ to test for numeric values
I've read the FAQ's on this, but they don't seem to answer
the question.
I have a variable that could contain any value( alpha,
alpha-numeric, or
numeric). If the value is NOT numeric, I need to change the
variables' value
to 0( as in zero ).
Examples:
$txtype=2.314; # is numeric, so keep the value
$txtype=-2.314; # is numeric, so keep the value
$txtype=7; # is numeric, so keep the value
$txtype=-7; # is numeric, so keep the value
$txtype=2.31.4; # is not numeric, so change the value to 0
$txtype=7-7; # is not numeric, so change the value to 0
$txtype=UNKNOWN; # is not numeric, so change the value to 0
$txtype=7+E09; # is not numeric( even though it really is
), so change
the value to 0
My guess is that I need a regex that will match on any
character that is:
not 0-9
or
more than one .
or
more than one -
or if - is not the first character of the string
Any ideas? Is it possible to do without using additional modules?
Thanks!
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RE: Regular expression to test for numeric values
How about this (direct from The Perl Cookbook[1]) ?? warn has nondigitsif /\D/; warn not a natural number unless /^\d+$/; # rejects -3 warn not an integer unless /^-?\d+$/; # rejects +3 warn not an integer unless /^[+-]?\d+$/; warn not a decimal number unless /^-?\d+\.?\d*$/; # rejects .2 warn not a decimal number unless /^-?(?:\d+(?:\.\d*)?|\.\d+)$/; warn not a C float unless /^([+-]?)(?=\d|\.\d)\d*(\.\d*)?([Ee]([+-]?\d+))?$/; The cookbook is the first place I look for this kind of stuff. HTH, Joe Dial [1] Perl Cookbook Tips and Tricks for Perl Programmers By Tom Christiansen, Nathan Torkington 1st Edition August 1998 www.oreilly.com ISBN: 1-56592-243-3 Current Version is: Perl Cookbook, 2nd Edition By Tom Christiansen, Nathan Torkington 2nd Edition August 2003 ISBN: 0-596-00313-7 964 pages, $49.95 US, $77.95 CA, £35.50 UK -Original Message- From: Motter, Jeffrey D [mailto:[EMAIL PROTECTED] Sent: Thursday, April 01, 2004 1:40 PM To: Perl-Win32-Users Subject: Regular expression to test for numeric values I've read the FAQ's on this, but they don't seem to answer the question. I have a variable that could contain any value( alpha, alpha-numeric, or numeric). If the value is NOT numeric, I need to change the variables' value to 0( as in zero ). Examples: $txtype=2.314; # is numeric, so keep the value $txtype=-2.314; # is numeric, so keep the value $txtype=7; # is numeric, so keep the value $txtype=-7; # is numeric, so keep the value $txtype=2.31.4; # is not numeric, so change the value to 0 $txtype=7-7; # is not numeric, so change the value to 0 $txtype=UNKNOWN; # is not numeric, so change the value to 0 $txtype=7+E09; # is not numeric( even though it really is ), so change the value to 0 My guess is that I need a regex that will match on any character that is: not 0-9 or more than one . or more than one - or if - is not the first character of the string Any ideas? Is it possible to do without using additional modules? Thanks! ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: regular expression question
Wagner, David --- Senior Programmer Analyst --- WGO wrote: Hey guys - what's with the HTML ? From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Wednesday, March 31, 2004 16:49 To: [EMAIL PROTECTED] Subject: regular expression question I have a $a which $a = this is a test; How can I change $a like: $a = this_is_a_test; No problem using $a, but if I were you, I would not. In the Perl sort $a, $b are the default variables to use and I would stay away if I were you. That said, $a =~ s/ /_/g; You could also use tr (slightly faster) : $a =~ tr/ /_/; -- ,-/- __ _ _ $Bill LuebkertMailto:[EMAIL PROTECTED] (_/ / )// // DBE CollectiblesMailto:[EMAIL PROTECTED] / ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/ -/-' /___/__/_/_http://dbecoll.tripod.com/ (My Perl/Lakers stuff) ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: regular expression replacement
Hi, I would like to know how can I replace the value c:\qqq\www\ to c:/qqq/www/. I tried several ways, but didn't manage to. Thanks, Eran s!\\!/!g ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: regular expression replacement
$Value = 'c:\qqq\www\'; $Value =~ s/\\/\//g; print $Value ; De: [EMAIL PROTECTED] em 28/11/2003 09:15 GMT Para: 'Kaufman Eran (StarHome)' [EMAIL PROTECTED] [EMAIL PROTECTED] cc: Assunto: RE: regular expression replacement Hi, I would like to know how can I replace the value c:\qqq\www\ to c:/qqq/www/. I tried several ways, but didn't manage to. Thanks, Eran s!\\!/!g ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs Tamanho do Documento: 3,78 KbytesLimite = 1200 Kbytes ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular Expression problem?
On Fri, 26 Sep 2003, Xu, Qiang (XSSC SGP) wrote: Ted S. wrote: Beckett Richard-qswi266 graced perl with these words of wisdom: That should have been s/.*\/// Don't you have to escape the period, too? s/\.*\/// No, we shouldn't, because here . stands for any single character except a new line. Well you are correct that a . regex character will match a period but consider if the matched string was 'aaabcd.txt' the unescaped period followed by * would match the string 'aaa'. Technically you are correct, you don't have to escape the . character unless of course you want the match to work correctly. [EMAIL PROTECTED] Carl Jolley All opinions are my own and not necessarily those of my employer ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular Expression problem?
Ted S. wrote: Beckett Richard-qswi266 graced perl with these words of wisdom: That should have been s/.*\/// Don't you have to escape the period, too? s/\.*\/// No, we shouldn't, because here . stands for any single character except a new line. thx, Regards, Xu Qiang ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular Expression problem?
On approximately 9/25/2003 6:45 AM, came the following characters from the keyboard of Xu, Qiang (XSSC SGP): Hi, all: I have a regular expression that I can't understand. Suppose $filename is a file name that includes the full path. Say, it is /u/scan/abc.jpg, The regular expression is: $filename =~ s|.*/||; I only know regular expressions such as s/.../.../, but this has the form of s|...|...| which I am confuse with. Anyone can explain the above regular expression as detailed as possible? Any delimiter can be used (other posters were correct about that). The delimiter used affects which delimiter character would need to be escaped in the regular expression. Generally, if something other than / is used as the delimiter character, it is chosen because it doesn't appear in the regular expression, so that there is no need to escape the delimiter character. Some exceptions to that rule include delimiter characters that add additional semantics, such as '. The expression in this case is stripping off leading directory names, probably hoping to leave an unqualified file name in the $filename variable. (Other posters claimed it was stripping the extension, but that is incorrect.) -- Glenn -- http://nevcal.com/ === Like almost everyone, I receive a lot of spam every day, much of it offering to help me get out of debt or get rich quick. It's ridiculous. -- Bill Gates ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular Expression problem?
Strohmeier Ruediger wrote: Hi Xu Qiang, unlike e.g. awk, vi or the shell, perl support different delimiter for regexes. When a slash is part of the regex or the substitution pattern, they can either be escaped (i.e. \/) or other characters can be used as delimiters. Thus the regex is equivalent to s/.*\/// and cuts away everything including the last slash changing /u/scan/abc.jpg into abc.jpg Yes, this is the correct result, thank you. Regards, Xu Qiang ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular Expression problem?
Glenn Linderman wrote: Any delimiter can be used (other posters were correct about that). The delimiter used affects which delimiter character would need to be escaped in the regular expression. Generally, if something other than / is used as the delimiter character, it is chosen because it doesn't appear in the regular expression, so that there is no need to escape the delimiter character. Some exceptions to that rule include delimiter characters that add additional semantics, such as '. The expression in this case is stripping off leading directory names, probably hoping to leave an unqualified file name in the $filename variable. (Other posters claimed it was stripping the extension, but that is incorrect.) Thank you, Glenn, you are right. The book said the purpose is to remove the path and keep the base name. I just couldn't understand the grammar in the expression, Now I can. Thanks for everyone who helped me. :) Regards, Xu Qiang ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Perldoc problem was Re: regular expression on military time
On Saturday, July 26, 2003 12:05 AM AEST, Ted S. wrote: On 25 Jul 2003, John McMahon wrote in perl: Ted When you produced the output of 'set' below how did you get to the CLI console (command line interpreter aka DOS prompt)? This console was opened in the 'Windows' directory. What was different in *HOW* you got to this console *TO HOW* you got to the console where you invoked 'perldoc.bat' in your earlier message (other than the directory they were opened in)? This console was opened in the 'Perl' directory. I'm not certain I understand your question? In all cases, I'm opening the console via the same shortcut that I've got in the Start menu. I then use the cd command to change directories if need be. And the console is always opened in the Windows directory. I'm just trying to get a picture of what you are doing and how you are doing it to see if I can suggest alternatives that might work. I have changed the subject as it was getting off topic. I assume you have installed Perl in the default location 'c:\perl' and your perl executable directory will be 'c:\perl\bin'. Since you use the same shortcut all the time and your path variable (as shown previously by your 'set' output 'PATH=C:\WINDOWS;C:\WINDOWS\COMMAND') does not include your Perl directory, to run scripts from the CLI console you would have to run them from the 'c:\perl\bin' directory. Ahh! I think I see the problem. Your previous postings: snip C:\Perlperldoc Usage: perldoc.bat [-h] [-r] [-i] [-v] [-t] [-u] [-m] [-n program] [-l] [- F] [-X ] PageName|ModuleName|ProgramName perldoc.bat -f PerlFunc perldoc.bat -q FAQKeywords The -h option prints more help. Also try perldoc perldoc to get acquainted with the system. /snip and snip C:\Perlperldoc perltoc Can't spawn command.com: No such file or directory at C:\PERL\BIN/perldoc.bat line 383. Can't spawn command.com: No such file or directory at C:\PERL\BIN/perldoc.bat line 383. Can't spawn command.com: No such file or directory at C:\PERL\BIN/perldoc.bat line 383. /snip shows you invoking perldoc in the 'c:\Perl' directory. Information provided so far suggests that this should not work at all, however it did, just not properly. I think this is probably because you don't have the perl executable directory (c:\perl\bin) in your path. If you were to repeat the exercise above in the 'c:\perl\bin' directory, I think it should work. If it does work, invoke 'path c:\perl\bin;%path%' at the prompt to change the path variable for the current console session then 'cd' to another directory and test again. It should work. If this is the answer, then you need to put 'c:\perl\bin' in your path, you probably should anyway. HTH John -- Regards John McMahon (mailto:[EMAIL PROTECTED]) Tired of Outlook Express/Outlook's messy quoting? Check out OE-Quotefix/Outlook-Quotefix via http://flash.to/oblivion ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Perldoc problem was Re: regular expression on military time
Just a quick note: - Original Message - From: John [EMAIL PROTECTED] To: Ted S. [EMAIL PROTECTED] Cc: Perl-Win32-Users [EMAIL PROTECTED] Sent: Sunday, July 27, 2003 3:07 AM Subject: Perldoc problem was Re: regular expression on military time clip If it does work, invoke 'path c:\perl\bin;%path%' at the prompt to change the path variable for the current console session then 'cd' to another directory and test again. It should work. If I remember right, this %path% syntax from the command prompt doesn't work under command.com. To change the path like that you'll need to repeat the current value and include the new entry on the end: path c:\windows;c:\windows\command;c:\perl\bin If this is the answer, then you need to put 'c:\perl\bin' in your path, you probably should anyway. HTH John -- Regards John McMahon (mailto:[EMAIL PROTECTED]) Tired of Outlook Express/Outlook's messy quoting? Check out OE-Quotefix/Outlook-Quotefix via http://flash.to/oblivion ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Perldoc problem was Re: regular expression on military time
On 27 Jul 2003, Gerry Green wrote in perl: Just a quick note: - Original Message - From: John [EMAIL PROTECTED] To: Ted S. [EMAIL PROTECTED] Cc: Perl-Win32-Users [EMAIL PROTECTED] Sent: Sunday, July 27, 2003 3:07 AM Subject: Perldoc problem was Re: regular expression on military time clip If it does work, invoke 'path c:\perl\bin;%path%' at the prompt to change the path variable for the current console session then 'cd' to another directory and test again. It should work. If I remember right, this %path% syntax from the command prompt doesn't work under command.com. To change the path like that you'll need to repeat the current value and include the new entry on the end: path c:\windows;c:\windows\command;c:\perl\bin This doesn't stick when I exit the DOS console. No, I don't know how to change environment variables or other such fun stuff. And heaven knows the help files/documentation are even more meager than what comes with most Perl modules! -- Ted Schuerzinger Homer Simpson: I'm sorry Marge, but sometimes I think we're the worst family in town. Marge: Maybe we should move to a larger community. http://www.snpp.com/episodes/7G04.html ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Perldoc problem was Re: regular expression on military time
On Monday, July 28, 2003 1:32 AM AEST, Gerry Green wrote: Just a quick note: - Original Message - From: John [EMAIL PROTECTED] To: Ted S. [EMAIL PROTECTED] Cc: Perl-Win32-Users [EMAIL PROTECTED] Sent: Sunday, July 27, 2003 3:07 AM Subject: Perldoc problem was Re: regular expression on military time clip If it does work, invoke 'path c:\perl\bin;%path%' at the prompt to change the path variable for the current console session then 'cd' to another directory and test again. It should work. If I remember right, this %path% syntax from the command prompt doesn't work under command.com. To change the path like that you'll need to repeat the current value and include the new entry on the end: path c:\windows;c:\windows\command;c:\perl\bin I am pretty sure it does, I have done that sort of thing for years (back to DOS 2.1), it certainly works in batch files, but I am not going to reboot into W98 just to test it. If this is the answer, then you need to put 'c:\perl\bin' in your path, you probably should anyway. HTH John -- Regards John McMahon (mailto:[EMAIL PROTECTED]) Tired of Outlook Express/Outlook's messy quoting? Check out OE-Quotefix/Outlook-Quotefix via http://flash.to/oblivion ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: regular expression on military time
On 25 Jul 2003, John McMahon wrote in perl: Ted When you produced the output of 'set' below how did you get to the CLI console (command line interpreter aka DOS prompt)? This console was opened in the 'Windows' directory. What was different in *HOW* you got to this console *TO HOW* you got to the console where you invoked 'perldoc.bat' in your earlier message (other than the directory they were opened in)? This console was opened in the 'Perl' directory. I'm not certain I understand your question? In all cases, I'm opening the console via the same shortcut that I've got in the Start menu. I then use the cd command to change directories if need be. And the console is always opened in the Windows directory. -- Ted Schuerzinger Homer Simpson: I'm sorry Marge, but sometimes I think we're the worst family in town. Marge: Maybe we should move to a larger community. http://www.snpp.com/episodes/7G04.html ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: regular expression on military time
On Fri, 18 Jul 2003, Ted S. wrote:
On 18 Jul 2003, Carl Jolley wrote in perl:
On Thu, 17 Jul 2003, Ted S. wrote:
On 17 Jul 2003, Tobias Hoellrich wrote in perl:
my @t=(08:00, 23:59, 00:00, aa:00, 24:00, 00:01,
8:00, 08.00, 36:12, 08:61 ); foreach(@t) {
unless(/^(\d{2}):(\d{2})$/ $124 $260) {
Forgive my ignorance, since I only use perl for basic things and
haven't yet gotten to text-munging:
I don't see what in your regex is capturing a $1 and a $2. (Yes, I
know what $1 and $2 are for. You can all stop laughing now. ;-)
The first set of parens maps to $1, the second to $2, etc.
Thanks, Carl and Bill. Wouldn't you know that this information about
parentheses mapping to $1 etc. actually *is* in the mess of documents that
is the Perl manpages. :-) (Specifically, perlre, but who can remember
what all the different page names actually stand for?)
You don't have to remember each of the perl pod file names. Just
remenber one command: perldoc perltoc.
[EMAIL PROTECTED] Carl Jolley
All opinions are my own and not necessarily those of my employer
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Re: regular expression question
[EMAIL PROTECTED] schrieb: I have a txt file like the bottom (I use cleardiff to compare 2 files and get this file): I would recommend that you look at Parse::RecDescent. Go grab the distribution from CPAN and have a look at the tutorial - the first example shows you how to parse a diff. http://theoryx5.uwinnipeg.ca/mod_perl/cpan-search?site=ftp.funet.fi;join=and;stem=no;arrange=file;case=clike;download=auto;age=;distinfo=3453 HTH, Thomas ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: regular expression question
The easiest way to reference all but the first character of a string is
probably:
$rest=substr $var,1;
You could also use a regex. Maybe search for a line that starts with and
then return 0 or more characters matching anything after that:
$var=~/^(.*)/;
$rest=$1;
You could also use rindex and select all the characters but one. Basically,
there are a lot of possibilities... a weekend with the Camel book or perldoc
would probably be rather informative.
Chris
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Friday, March 28, 2003 11:58 AM
To: [EMAIL PROTECTED]
Subject: RE: regular expression question
I have another question,
I have string like GENERATION 116, How can I get rid of
, of the
string?
Thanks
Lixin
-Original Message-
From: Todd Hayward [mailto:[EMAIL PROTECTED]
Sent: Friday, March 28, 2003 11:30 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: regular expression question
load the file into a file handle, and iterate through it line
by line. Pipe
each line to the file handle (the | is not a typo in the example)...
So,
open(fHandle, myfile.txt|);
while (defined ($line = fHandle)){
if ($line ~= /generic/ig){
do something}
};
Hope this helps,
NuTs
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Friday, March 28, 2003 9:40 AM
Subject: regular expression question
Dear all
I have a txt file like the bottom (I use cleardiff to
compare 2 files and
get this file):
I will check whether the line includes
___GENERIC_MULTILINE___ or not,
if
it includes, I will get the following information
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
without sign,
How can I do it?
Thanks a lot in advance!
Lixin
#
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
24c24,30
___GENERIC_MULTILINE___
---
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
##
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RE: regular expression question
I have another question, I have string like GENERATION 116, How can I get rid of , of the string? If you want to remove the 1st character if it is a , then use this... $var =~ s///; $var =~ s/^//; #This removes it only if it is the first character ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: regular expression question
This works:
open (INPUT ,your_input.txt) || die $!;
open (OUTPUT, your_output.txt) || die $!;
while (INPUT) {
if (m/GENERIC_MULTILINE/) {
$_=GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116\n;}
print OUTPUT $_;}
close (OUTPUT);
close (INPUT);
viktoras
[EMAIL PROTECTED] wrote:
Dear all
I have a txt file like the bottom (I use cleardiff to compare 2 files and
get this file):
I will check whether the line includes ___GENERIC_MULTILINE___ or not, if
it includes, I will get the following information
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
without sign,
How can I do it?
Thanks a lot in advance!
Lixin
#
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
24c24,30
___GENERIC_MULTILINE___
---
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
##
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RE: regular expression question
Thanks a lot for your great help, another question,
if I want to switch all \n to \\n in a string how can I do it?
Like
$var = This is a book\nThat is a desk\nwho is that man?\n;
I did like
$var =~ s/\n/\\n/g;
but it does nt work. What is wrong with it?
Thanks
Lixin
-Original Message-
From: Todd Hayward [mailto:[EMAIL PROTECTED]
Sent: Friday, March 28, 2003 12:26 PM
To: [EMAIL PROTECTED]
Subject: Re: regular expression question
Borrowing from the previous example of:
open(fHandle, myfile.txt|);
while (defined ($line = fHandle)){
if ($line ~= /generic/ig){
do something}
};
Add this line: $line =~ s/\//g;
This is what your code block should look like:
open(fHandle, myfile.txt|);
while (defined ($line = fHandle)){
$line =~ s/\//g;
if ($line =~ /generic/ig){
do something}
};
The 3rd line will effectively delete any instance of that it finds. If
you want it to only remove the first instance of from each line, remove
the g from the end. This makes it global to the string, thus
substituting no character for each it finds.
HTH,
NuTs
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Friday, March 28, 2003 10:58 AM
Subject: RE: regular expression question
I have another question,
I have string like GENERATION 116, How can I get rid of , of the
string?
Thanks
Lixin
-Original Message-
From: Todd Hayward [mailto:[EMAIL PROTECTED]
Sent: Friday, March 28, 2003 11:30 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: regular expression question
load the file into a file handle, and iterate through it line by line.
Pipe
each line to the file handle (the | is not a typo in the example)...
So,
open(fHandle, myfile.txt|);
while (defined ($line = fHandle)){
if ($line ~= /generic/ig){
do something}
};
Hope this helps,
NuTs
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Friday, March 28, 2003 9:40 AM
Subject: regular expression question
Dear all
I have a txt file like the bottom (I use cleardiff to compare 2 files
and
get this file):
I will check whether the line includes ___GENERIC_MULTILINE___ or not,
if
it includes, I will get the following information
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
without sign,
How can I do it?
Thanks a lot in advance!
Lixin
#
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
24c24,30
___GENERIC_MULTILINE___
---
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
##
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RE: regular expression question
I have another question,
I have string like GENERATION 116, How can I get rid of , of the
string?
Thanks
Lixin
-Original Message-
From: Todd Hayward [mailto:[EMAIL PROTECTED]
Sent: Friday, March 28, 2003 11:30 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: Re: regular expression question
load the file into a file handle, and iterate through it line by line. Pipe
each line to the file handle (the | is not a typo in the example)...
So,
open(fHandle, myfile.txt|);
while (defined ($line = fHandle)){
if ($line ~= /generic/ig){
do something}
};
Hope this helps,
NuTs
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Friday, March 28, 2003 9:40 AM
Subject: regular expression question
Dear all
I have a txt file like the bottom (I use cleardiff to compare 2 files and
get this file):
I will check whether the line includes ___GENERIC_MULTILINE___ or not,
if
it includes, I will get the following information
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
without sign,
How can I do it?
Thanks a lot in advance!
Lixin
#
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
24c24,30
___GENERIC_MULTILINE___
---
GENERATION 116
# Impossible dependency
# Needed to prevent FC4700 to CX-series upgrades
DEPEND Navisphere 2.0.0.0.0
DEPEND Navisphere 1.0.0.0.0
GENDEPEND Navisphere 116
3,4c3,4
REVISION ___INTERNAL_REV___
DISPLAYREV ___EXTERNAL_REV___
---
REVISION 02041405.002
DISPLAYREV 02.04.1.40.5.002
##
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Re: Regular Expression matching problem
On Sun, 9 Mar 2003, Electron One wrote:
Hello Everyone,
If I have a file that contains this,
test3.txt##
wilma
wimagren was here
twilma was type wilma
wilma
wilma
wilma
twowilmase
##
and i have a perl script that contains this,
###perlname.pl
#!/usr/bin/perl
while(){
chomp;
if(/wilma\s+/){
print wilma was mentioned\n;
}
}
#
and I type, perl -w perlname.pl test3.txt
shouldnt the output be, wilma was mentioned \n, 7 times? My output file
only mentions it once. What am I doing wrong?
Two things:
One since your are chomping the input line, your regex which requires
trailing white space after 'wilma' wont't match those lines which end with
the characters wilma. I suggest you change your regex to: /wilma\b/.
Alternatively you could just take out the chomp.
Two you are only checking for one instance of your regex per input line.
I suggest you change your code from an 'if' to:
while(/wilma\b/g) {
You should get 6 matches on your input file not seven.
[EMAIL PROTECTED] Carl Jolley
All opinions are my own and not necessarily those of my employer
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Re: Regular Expression matching problem
Not if the fourth to sixth lines don't have a space before the end of line.
Try if(/wilma\s*/). Actually, you'll need to use if(/wilma/g) to catch them
all.
Phil.
|-+---
| | Electron One|
| | [EMAIL PROTECTED]|
| | Sent by:|
| | [EMAIL PROTECTED]|
| | veState.com |
| | |
| | |
| | 10/03/03 02:29 |
| | |
|-+---
--|
|
|
| To: [EMAIL PROTECTED]
|
| cc: electron One [EMAIL PROTECTED]
|
| Subject: Regular Expression matching problem
|
--|
Hello Everyone,
If I have a file that contains this,
test3.txt##
wilma
wimagren was here
twilma was type wilma
wilma
wilma
wilma
twowilmase
##
and i have a perl script that contains this,
###perlname.pl
#!/usr/bin/perl
while(){
chomp;
if(/wilma\s+/){
print wilma was mentioned\n;
}
}
#
and I type, perl -w perlname.pl test3.txt
shouldnt the output be, wilma was mentioned \n, 7 times? My output file
only mentions it once. What am I doing wrong?
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Re: Regular Expression matching problem
Electron One [EMAIL PROTECTED] wrote:
while(){
chomp;
if(/wilma\s+/){
print wilma was mentioned\n;
}
}
#
and I type, perl -w perlname.pl test3.txt
shouldnt the output be, wilma was mentioned \n, 7 times? My
output file only mentions it once. What am I doing wrong?
What do you think the '\s+' means in that regular expression?
After you've chomped them, only one line has whitespace after
'wilma', so only one matches.
--
Keith C. Ivey [EMAIL PROTECTED]
Washington, DC
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Re: regular expression question
On Fri, 22 Nov 2002 [EMAIL PROTECTED] wrote:
all,
I want to check the first line of the file if it is machine or not, like
The first line of the file is:
Job \nest and \toolbox VOBs began execution on 9/6/02 at 2:00:11 AM.
my code is like:
if (!-z $file)
{
open(LOG_FILE, $file) or warn can not open $file:$!\n;
my @read_lines = LOG_FILE;
close (LOG_FILE);
next unless chomp($read_lines[0]) =~ m#\\nest and \toolbox VOBs\#;
}
it did not work for regular expression, can you help me to figure what is
wrong with it?
You are performing your regex match on the result of the function call:
chomp($read_lines[0]). The value of chomp will be 1 if it removed a
new-line character from the end of $read_lines[0] and it will be zero if
it didn't remove a new-line character, i.e. if the last character was not
the new-line character (actually the $/ string which has a default value
of \n). Somehow I don't believe that your regex will match either a 1
or a 0.
However, even if you anchor the regex on the variable $read_lines[0],
I don't believe that the regex match string is correct considering
that fact that it contains the representation of a new-line character (the
\n) and a tab (the \t). Also note that it is not necessary to escape the
double quote characters inside the regex match string unless the quote is
used to delimit the match string (you used # to delimit the regex match
string, my code below uses the default regex delimter, /) I believe the
the following will fix your problem as far as the regex is concerned:
next unless $read_lines[0] =~ /\\nest and \\toolbox VOBs/;
Take out the chomp, it serves no purpose based on the code fragment
you showed. Also reconsider your logic. Exactly what do you expect to
happen if the regex doesn't match, i.e. if the 'next' is done what
code will then execute next?
[EMAIL PROTECTED] Carl Jolley
All opinions are my own and not necessarily those of my employer
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RE: regular expression question
On Fri, 22 Nov 2002, Stovall, Adrian M. wrote:
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Friday, November 22, 2002 4:38 PM
To: Stovall, Adrian M.; [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: regular expression question
I tried that, it does not work for me!
Lixin
This code should print yes, if it does, then the code works for you...
$line = 'Job \nest and \toolbox VOBs began execution on 9/6/02 at
2:00:11 AM.';
print $line\n;
if ($line =~ m#\\\nest and \\toolbox VOBs\#) { print yes; }
else { print no; }
Your code may work fine but as long as his code is archoring
the regex to the result of the function call: chomp($read_lines[0])
it won't. Unless of course he changes the regex match string to:
m#^[01]$#;
Note that the result of doing chomp on a scalar will be one of
two values, 0 or 1.
[EMAIL PROTECTED] Carl Jolley
All opinions are my own and not necessarily those of my employer
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RE: regular expression question
Sorry, I did not state quite clear, if it is machine or not, I want to say
if I it is the right file or not...
Lixin
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Friday, November 22, 2002 5:07 PM
Cc: [EMAIL PROTECTED]
Subject: regular expression question
all,
I want to check the first line of the file if it is machine or not, like
The first line of the file is:
Job \nest and \toolbox VOBs began execution on 9/6/02 at 2:00:11 AM.
my code is like:
if (!-z $file)
{
open(LOG_FILE, $file) or warn can not open $file:$!\n;
my @read_lines = LOG_FILE;
close (LOG_FILE);
next unless chomp($read_lines[0]) =~ m#\\nest and \toolbox VOBs\#;
}
it did not work for regular expression, can you help me to figure what is
wrong with it?
Thanks a lot!
Lixin
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RE: regular expression question
Cai Lixin said:
all,
I want to check the first line of the file if it is machine
or not, like
The first line of the file is:
Job \nest and \toolbox VOBs began execution on 9/6/02 at 2:00:11 AM.
my code is like:
if (!-z $file)
{
open(LOG_FILE, $file) or warn can not open $file:$!\n;
my @read_lines = LOG_FILE;
close (LOG_FILE);
next unless chomp($read_lines[0]) =~ m#\\nest and
\toolbox VOBs\#;
}
it did not work for regular expression, can you help me to
figure what is wrong with it?
You forgot to escape the backslashes...change your code to read:
next unless chomp($read_lines[0]) =~ m#\\\nest and \\toolbox VOBs\#)
perl -e sub Sub{return reverse(@_);}$_='.$yyye k ca i Xl $yyye jX $yyye
hto ZfX tq $uQ';s+[ \$]++g;s-j-P-;s^yyy^r^g;s:i:H:;s!X!
!g;s|Z|n|;s*Q*J*;s{q}{s}g;s(f)(A);$print=join('',Sub(split('')));system(
'echo',$print);
To the optimist, the glass is half full.
To the pessimist, the glass is half empty.
To the engineer, the glass is twice as big as it needs to be.
Adrian Okay, I won't top-post unless it's an emergency Stovall
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RE: regular expression question
I tried that, it does not work for me!
Lixin
-Original Message-
From: Stovall, Adrian M. [mailto:[EMAIL PROTECTED]]
Sent: Friday, November 22, 2002 5:28 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: regular expression question
Cai Lixin said:
all,
I want to check the first line of the file if it is machine
or not, like
The first line of the file is:
Job \nest and \toolbox VOBs began execution on 9/6/02 at 2:00:11 AM.
my code is like:
if (!-z $file)
{
open(LOG_FILE, $file) or warn can not open $file:$!\n;
my @read_lines = LOG_FILE;
close (LOG_FILE);
next unless chomp($read_lines[0]) =~ m#\\nest and
\toolbox VOBs\#;
}
it did not work for regular expression, can you help me to
figure what is wrong with it?
You forgot to escape the backslashes...change your code to read:
next unless chomp($read_lines[0]) =~ m#\\\nest and \\toolbox VOBs\#)
perl -e sub Sub{return reverse(@_);}$_='.$yyye k ca i Xl $yyye jX $yyye
hto ZfX tq $uQ';s+[ \$]++g;s-j-P-;s^yyy^r^g;s:i:H:;s!X!
!g;s|Z|n|;s*Q*J*;s{q}{s}g;s(f)(A);$print=join('',Sub(split('')));system(
'echo',$print);
To the optimist, the glass is half full.
To the pessimist, the glass is half empty.
To the engineer, the glass is twice as big as it needs to be.
Adrian Okay, I won't top-post unless it's an emergency Stovall
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RE: regular expression question
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Friday, November 22, 2002 4:38 PM
To: Stovall, Adrian M.; [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: regular expression question
I tried that, it does not work for me!
Lixin
This code should print yes, if it does, then the code works for you...
$line = 'Job \nest and \toolbox VOBs began execution on 9/6/02 at
2:00:11 AM.';
print $line\n;
if ($line =~ m#\\\nest and \\toolbox VOBs\#) { print yes; }
else { print no; }
perl -e sub Sub{return reverse(@_);}$i='ohce';$_='.$yyye k ca i Xl
$yyye jX $yyyehto ZfX tq $uQ';s+[ \$]++g;s-j-P-;s^yyy^r^g;s:i:H:;s!X!
!g;s|Z|n|;s*Q*J*;s{q}{s}g;s(f)(A);system(join('',Sub(split('',$i))),(joi
n('',Sub(split('');
To the optimist, the glass is half full.
To the pessimist, the glass is half empty.
To the engineer, the glass is twice as big as it needs to be.
Adrian Okay, I won't top-post unless it's an emergency Stovall
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RE: regular expression question
Yes, it works fine for me!
Thanks a lot!
Have a nice day.
Lixin
-Original Message-
From: Stovall, Adrian M. [mailto:[EMAIL PROTECTED]]
Sent: Friday, November 22, 2002 5:42 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: regular expression question
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Friday, November 22, 2002 4:38 PM
To: Stovall, Adrian M.; [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: RE: regular expression question
I tried that, it does not work for me!
Lixin
This code should print yes, if it does, then the code works for you...
$line = 'Job \nest and \toolbox VOBs began execution on 9/6/02 at
2:00:11 AM.';
print $line\n;
if ($line =~ m#\\\nest and \\toolbox VOBs\#) { print yes; }
else { print no; }
perl -e sub Sub{return reverse(@_);}$i='ohce';$_='.$yyye k ca i Xl
$yyye jX $yyyehto ZfX tq $uQ';s+[ \$]++g;s-j-P-;s^yyy^r^g;s:i:H:;s!X!
!g;s|Z|n|;s*Q*J*;s{q}{s}g;s(f)(A);system(join('',Sub(split('',$i))),(joi
n('',Sub(split('');
To the optimist, the glass is half full.
To the pessimist, the glass is half empty.
To the engineer, the glass is twice as big as it needs to be.
Adrian Okay, I won't top-post unless it's an emergency Stovall
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RE: Regular Expression Problem
Lee Cullip wrote, on Wednesday, November 20, 2002 06:50
: Can anybody tell me what is wrong with the following line ?
:
: $prmpt = /(^.*[:]\/home\/oracle[:=]{1,2})/;
:
: I'm trying to use the value of $prmpt in a call to Net::Telnet but before I can
: use the variable $prmpt, perl is complaining about the format of the pattern.
Please try to post in plain text. Thanks.
Do you really want $prmpt to be true if $_ matched your pattern,
and false if it didn't? Perhaps you meant =~ instead of =.
What exactly is Perl's complaint about your pattern? I have a
few suggestions:
o Use a different pattern delimiter if you want to match
forward slashes, to avoid the leaning toothpick
phenomenon.
o You don't need to turn a single character into a
character class: remove the brackets from the [:].
o Unless you need $1 later, you don't need parentheses
around the whole pattern. My guess is that Perl
didn't like having the start anchor inside parens.
Rewriting your pattern, then, I get this:
$prmpt =~ !^(.*:/home/oracle[:=]{1,2})!;
$prmpt = $1;
Is that what you really meant?
Good luck,
Joe
==
Joseph P. Discenza, Sr. Programmer/Analyst
mailto:[EMAIL PROTECTED]
Carleton Inc. http://www.carletoninc.com
574.243.6040 ext. 300fax: 574.243.6060
Providing Financial Solutions and Compliance for over 30 Years
* Please note that our Area Code has changed to 574! *
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Re: Regular Expression Problem
Thanks for replying joe,
maybe if you see a bit more code you can get an idea for what I'm trying to
do :
use Net::Telnet;
use IO::File;
$prmpt =~ /^(.*:\/home\/oracle[:=]{1,2})/;
$telnet = new Net::Telnet(-prompt = $prmpt, -Errmode = 'die');
$telnet-open($HOST);
$telnet-login($USER,$PASSWORD);
$telnet-cmd(-string =,-Timeout = 20,-prompt = $prmpt);
@lines = $telnet-cmd(cat .netrc);
$telnet-cmd(-string =,-Timeout = 20,-prompt = $prmpt);
$netrctxt = new IO::File;
$netrctxt-open('c:\temp\netrc.txt') or die Cannot open
C:\temp\netrc.txt for writing;
$netrctxt-print(@lines);
$netrctxt-close;
$telnet-close;
Don't know if you can see what I'm trying to do, but I want the telnet
prompt to match $prmpt. Maybe I'm Tackling it the wrong way, but I'm still
not a perl guru (yet ;-) ).
I still get the following error message though :
bad match operator: opening delimiter missing: ^(.*:/home/oracle[:=]{1,2})
at C:\Documents and Settings\lcullip\My
Documents\programming\perl\oracheck.pl line 41
WTH am I doing wrong ?
TIA
Lee
- Original Message -
From: Joseph P. Discenza [EMAIL PROTECTED]
To: Lee Cullip [EMAIL PROTECTED];
[EMAIL PROTECTED]
Sent: Wednesday, November 20, 2002 12:35 PM
Subject: RE: Regular Expression Problem
Lee Cullip wrote, on Wednesday, November 20, 2002 06:50
: Can anybody tell me what is wrong with the following line ?
:
: $prmpt = /(^.*[:]\/home\/oracle[:=]{1,2})/;
:
: I'm trying to use the value of $prmpt in a call to Net::Telnet but before
I can
: use the variable $prmpt, perl is complaining about the format of the
pattern.
Please try to post in plain text. Thanks.
Do you really want $prmpt to be true if $_ matched your pattern,
and false if it didn't? Perhaps you meant =~ instead of =.
What exactly is Perl's complaint about your pattern? I have a
few suggestions:
o Use a different pattern delimiter if you want to match
forward slashes, to avoid the leaning toothpick
phenomenon.
o You don't need to turn a single character into a
character class: remove the brackets from the [:].
o Unless you need $1 later, you don't need parentheses
around the whole pattern. My guess is that Perl
didn't like having the start anchor inside parens.
Rewriting your pattern, then, I get this:
$prmpt =~ !^(.*:/home/oracle[:=]{1,2})!;
$prmpt = $1;
Is that what you really meant?
Good luck,
Joe
==
Joseph P. Discenza, Sr. Programmer/Analyst
mailto:[EMAIL PROTECTED]
Carleton Inc. http://www.carletoninc.com
574.243.6040 ext. 300fax: 574.243.6060
Providing Financial Solutions and Compliance for over 30 Years
* Please note that our Area Code has changed to 574! *
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Re: Regular Expression Problem
On 20/11/2002 13:06:29 Lee Cullip wrote:
Thanks for replying Joe,
maybe if you see a bit more code you can get an idea for what I'm trying
to
do :
use Net::Telnet;
use IO::File;
$prmpt =~ /^(.*:\/home\/oracle[:=]{1,2})/;
[snip]
I still get the following error message though :
bad match operator: opening delimiter missing: ^(.*:/home/oracle[:
=]{1,2})
at C:\Documents and Settings\lcullip\My
Documents\programming\perl\oracheck.pl line 41
WTH am I doing wrong ?
You are not giving us the exact script, perhaps because you are re-typing
instead of copypaste.
I pasted the above regex into
perl -pwe s/^(.*:\/home\/oracle[:=]{1,2})/$1/
and it duly replaced :/home/oracle: with :/home/oracle:
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Sorry (was RE: Regular Expression Help)
My fault: I'm sorry. I avoid usenet for the same reasons.
I apologise.
But really, such use of English makes me wonder about
possible use of Perl.
However, I must say that it is not the whole group, just
me.
Lee
At 04:19 12/06/2002, Allegakoen, Justin Devanandan wrote:
Lets not point fingers here, but this group is starting to become like
comp.lang.perl.misc
I subscribed to this list to avoid that type of . . . [insert disparaging
term here] . . .
We have all sort of people inside this list, from many country, some of our
grammer are
worst then others, but lets remember that we is all brothers in alms with
our won common
language - Perl ; )
Flame at will!
-Original Message-
From: Ron Grabowski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 12, 2002 2:16 AM
To: [EMAIL PROTECTED]
Subject: Re: Regular Expression Help
phone number. The three digit is
are
digits
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Lee Goddard
perl -e while(1){print rand0.5?chr 47:chr 92}
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RE: Regular Expression Help
Lets not point fingers here, but this group is starting to become like comp.lang.perl.misc I subscribed to this list to avoid that type of . . . [insert disparaging term here] . . . We have all sort of people inside this list, from many country, some of our grammer are worst then others, but lets remember that we is all brothers in alms with our won common language - Perl ; ) Flame at will! -Original Message- From: Ron Grabowski [mailto:[EMAIL PROTECTED]] Sent: Wednesday, June 12, 2002 2:16 AM To: [EMAIL PROTECTED] Subject: Re: Regular Expression Help phone number. The three digit is are digits ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: Regular Expression Help
This worked for me,
$a = 12345678904539;
@numbers = split(/(\d{2})/, $a);
$NewA = join(' ', @numbers);
print \nNew A: $NewA;
hth,
Joe Y.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of
steve silvers
Sent: Monday, June 10, 2002 8:39 AM
To: [EMAIL PROTECTED]
Subject: Regular Expression Help
How can I put a white space between every second number.
I have $a = (12345678904539);
I want 12 34 56 78 90 45 39
I'm trying
$a =~ s/\\d[2*]/ /g; #This obviously dosen't work :-(
Also how can I tell if there are 3,4, or 5 digits.
$3dig = (888);
$4dig = ();
$5dig = (8);
Thanks in advance.
Steve.
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RE: Regular Expression Help
Joseph Youngquist wrote:
This worked for me,
$a = 12345678904539;
@numbers = split(/(\d{2})/, $a);
$NewA = join(' ', @numbers);
print \nNew A: $NewA;
Or, more simply, s/(\d\d)(?=\d\d)/$1 /g;
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RE: Regular Expression Help
On Mon, 10 Jun 2002 09:10:05
Joseph Youngquist wrote:
This worked for me,
$a = 12345678904539;
@numbers = split(/(\d{2})/, $a);
$NewA = join(' ', @numbers);
print \nNew A: $NewA;
snip
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of
steve silvers
How can I put a white space between every second number.
I have $a = (12345678904539);
I want 12 34 56 78 90 45 39
And this worked for me:
s%([\d]{2})%$1 %g;
---
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[EMAIL PROTECTED]
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Re: Regular Expression Help
On Mon, 10 Jun 2002 13:38:52
steve silvers wrote:
How can I put a white space between every second number.
I have $a = (12345678904539);
I want 12 34 56 78 90 45 39
I'm trying
$a =~ s/\\d[2*]/ /g; #This obviously dosen't work :-(
Also how can I tell if there are 3,4, or 5 digits.
$3dig = (888);
$4dig = ();
$5dig = (8);
Thanks in advance.
Steve.
Oops, forgot to answer the second part. But we really need more info.
Is each number (i.e. a run of digits) in a separate variable as you have shown? If
so simply find length($5dig), etc.
If it is not as simple as this (e.g. are the numbers in a file which you will read in,
is there one number per line, is there anything else on the line, do you have to find
all the numbers or just the first or just the last, etc.) something like
print 3 digits found: $\n if ($string =~ m%[\d]{3}%);
may get you started, but note that 3 digits will be found even when there are 4 or 5,
so do the tests in the right order.
---
Steve Martin
[EMAIL PROTECTED]
PS I have learned something from this post, I didn't know you could define a string
using brackets as you have done...
___
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on the big Island. Enter Now!
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RE: Regular Expression Help
PS I have learned something from this post, I didn't know you could
define a string using brackets as you have done...
Except that what you learned:
$a = (12345678904539);
is a bad practise. It is a novice mistake. What is being done here is
an anonymous list has been created with one element: 12345678904539 .
The scalar $a is then assigned the last element of the array (which
in this case is also the first element. To see this, try:
$a = (123, 456);
print $a;
# you will get: 456
Also, the value is not a string. It is a number. The following is an
error:
$a = (abc);
# this is an error!
--
Mike Arms
-Original Message-
From: Stephen J Martin [mailto:[EMAIL PROTECTED]]
Sent: Monday, June 10, 2002 8:51 AM
To: steve silvers
Cc: [EMAIL PROTECTED]
Subject: Re: Regular Expression Help
On Mon, 10 Jun 2002 13:38:52
steve silvers wrote:
How can I put a white space between every second number.
I have $a = (12345678904539);
I want 12 34 56 78 90 45 39
I'm trying
$a =~ s/\\d[2*]/ /g; #This obviously dosen't work :-(
Also how can I tell if there are 3,4, or 5 digits.
$3dig = (888);
$4dig = ();
$5dig = (8);
Thanks in advance.
Steve.
Oops, forgot to answer the second part. But we really need more info.
Is each number (i.e. a run of digits) in a separate variable as you have
shown? If so simply find length($5dig), etc.
If it is not as simple as this (e.g. are the numbers in a file which you
will read in, is there one number per line, is there anything else on the
line, do you have to find all the numbers or just the first or just the
last, etc.) something like
print 3 digits found: $\n if ($string =~ m%[\d]{3}%);
may get you started, but note that 3 digits will be found even when there
are 4 or 5, so do the tests in the right order.
---
Steve Martin
[EMAIL PROTECTED]
PS I have learned something from this post, I didn't know you could define a
string using brackets as you have done...
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RE: Regular Expression Help
--- Stephen J Martin [EMAIL PROTECTED] wrote:
On Mon, 10 Jun 2002 09:10:05
Joseph Youngquist wrote:
This worked for me,
$a = 12345678904539;
@numbers = split(/(\d{2})/, $a);
$NewA = join(' ', @numbers);
print \nNew A: $NewA;
snip
-Original Message-
From:
[EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On
Behalf Of
steve silvers
How can I put a white space between every second
number.
I have $a = (12345678904539);
I want 12 34 56 78 90 45 39
And this worked for me:
s%([\d]{2})%$1 %g;
---
Steve Martin
[EMAIL PROTECTED]
snip
When I run this snippet below, using the three
approaches from others, there are three slightly
different results. See results at end.
Please note spaces at start and end of some results.
Also, note results with a number of odd length.
Perhaps the difference is not significant to the OP.
I wish I could figure out a better way,
Jim
# -
#!perl
use strict;
use warnings;
if (1){
# try with several numbers of diff lengths.
my $a = 12345671234;
my $b = 123456712345;
my $c = 1234567123456;
print \n;
foreach my $n ( $a, $b, $c, ){
# Joseph
my @numbers = split(/(\d{2})/, $n);
print \t*, join(' ', @numbers), *\n;
# Larry
my $L = $n;
$L =~ s/(\d\d)(?=\d\d)/$1 /g;
print \t*$L* \n;
# Stephen
my $s = $n;
$s =~ s%([\d]{2})%$1 %g;
print \t*$s* \n\n;
}
} # end of if 0.
# -
produces:
* 12 34 56 71 23 4*
*12 34 56 71 234*
*12 34 56 71 23 4*
* 12 34 56 71 23 45*
*12 34 56 71 23 45*
*12 34 56 71 23 45 *
* 12 34 56 71 23 45 6*
*12 34 56 71 23 456*
*12 34 56 71 23 45 6*
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RE: Regular Expression Help
Jim Angstadt wrote:
When I run this snippet below, using the three
approaches from others, there are three slightly
different results. See results at end.
Please note spaces at start and end of some results.
Also, note results with a number of odd length.
Perhaps the difference is not significant to the OP.
I wish I could figure out a better way,
Jim
# -
#!perl
use strict;
use warnings;
if (1){
# try with several numbers of diff lengths.
my $a = 12345671234;
my $b = 123456712345;
my $c = 1234567123456;
print \n;
foreach my $n ( $a, $b, $c, ){
# Joseph
my @numbers = split(/(\d{2})/, $n);
print \t*, join(' ', @numbers), *\n;
# Larry
my $L = $n;
$L =~ s/(\d\d)(?=\d\d)/$1 /g;
print \t*$L* \n;
# Stephen
my $s = $n;
$s =~ s%([\d]{2})%$1 %g;
print \t*$s* \n\n;
}
} # end of if 0.
# -
produces:
* 12 34 56 71 23 4*
*12 34 56 71 234*
*12 34 56 71 23 4*
* 12 34 56 71 23 45*
*12 34 56 71 23 45*
*12 34 56 71 23 45 *
* 12 34 56 71 23 45 6*
*12 34 56 71 23 456*
*12 34 56 71 23 45 6*
Nice job of breaking down the problem.
If you tweak my regular expression ever-so-slightly, it does a much cleaner
job:
#!perl
use strict;
use warnings;
if (1){
# try with several numbers of diff lengths.
my $a = 12345671234;
my $b = 123456712345;
my $c = 1234567123456;
print \n;
foreach my $n ( $a, $b, $c, ){
# Joseph
my @numbers = split(/(\d{2})/, $n);
print \t*, join(' ', @numbers), *\n;
# Larry
my $L = $n;
$L =~ s/(\d\d)(?=\d)/$1 /g; # Note: only one digit in the look-ahead
print \t*$L* \n;
# Stephen
my $s = $n;
$s =~ s%([\d]{2})%$1 %g;
print \t*$s* \n\n;
}
} # end of if 0.
__END__
Output:
* 12 34 56 71 23 4*
*12 34 56 71 23 4*
*12 34 56 71 23 4*
* 12 34 56 71 23 45*
*12 34 56 71 23 45*
*12 34 56 71 23 45 *
* 12 34 56 71 23 45 6*
*12 34 56 71 23 45 6*
*12 34 56 71 23 45 6*
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RE: Regular Expression Help
On Mon, 10 Jun 2002 10:29:28 Arms, Mike wrote: PS I have learned something from this post, I didn't know you could define a string using brackets as you have done... Except that what you learned: $a = (12345678904539); is a bad practise. It is a novice mistake. What is being done here is snip Ah yes, of course, thanks for pointing that out. I had a feeling something was fishy there but didn't know what it was. Luckily I was not planning to use my new knowledge any time soon ... Also luckily my solution works with real strings too (I tested both, honestly). Steve. ___ WIN a first class trip to Hawaii. Live like the King of Rock and Roll on the big Island. Enter Now! http://r.lycos.com/r/sagel_mail/http://www.elvis.lycos.com/sweepstakes ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Regular Expression Help
Sorry for not elaborating on the 3 and 4 digit thing. I have a form that the
end user inters there phone number. The three digit is for the area code and
first 3 digits of the phone number. I just want to add some server side
validation besides just javascript. So I have something like:
$query-('areacode') =~ m/\d{3}/g;
I want to make sure that there is only 3 digits, not two, or four.
Steve.
From: Stephen J Martin [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: steve silvers [EMAIL PROTECTED]
CC: [EMAIL PROTECTED]
Subject: Re: Regular Expression Help
Date: Mon, 10 Jun 2002 14:50:53
On Mon, 10 Jun 2002 13:38:52
steve silvers wrote:
How can I put a white space between every second number.
I have $a = (12345678904539);
I want 12 34 56 78 90 45 39
I'm trying
$a =~ s/\\d[2*]/ /g; #This obviously dosen't work :-(
Also how can I tell if there are 3,4, or 5 digits.
$3dig = (888);
$4dig = ();
$5dig = (8);
Thanks in advance.
Steve.
Oops, forgot to answer the second part. But we really need more info.
Is each number (i.e. a run of digits) in a separate variable as you have
shown? If so simply find length($5dig), etc.
If it is not as simple as this (e.g. are the numbers in a file which you
will read in, is there one number per line, is there anything else on the
line, do you have to find all the numbers or just the first or just the
last, etc.) something like
print 3 digits found: $\n if ($string =~ m%[\d]{3}%);
may get you started, but note that 3 digits will be found even when there
are 4 or 5, so do the tests in the right order.
---
Steve Martin
[EMAIL PROTECTED]
PS I have learned something from this post, I didn't know you could define
a string using brackets as you have done...
___
WIN a first class trip to Hawaii. Live like the King of Rock and Roll
on the big Island. Enter Now!
http://r.lycos.com/r/sagel_mail/http://www.elvis.lycos.com/sweepstakes
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RE: Regular Expression
Hi Steve,
This may not be the most eficient way, but it seems to work.
Someone will undoubtedly offer another (probably better) way to do it.
hth
Toby
snip
my @numbers = qw(01 02 03 03 05 08 09 12 14 13 11 18 17 12 15 16 15 16 12 13
14 16 17 22 23 24 25 25 23 22 21 20);
my %counts;
occur_in_array(\@numbers);
sub occur_in_array
{
my $aRef = shift;
foreach (@$aRef)
{
if (!exists($counts{$_}))
{
$counts{$_} = 1;
}
else
{
$counts{$_}++;
}
}
}
while (($key,$value) = each %counts)
{
print $key = $value\n;
}
/snip
-Original Message-
From: steve silvers [mailto:[EMAIL PROTECTED]]
Sent: Monday, May 06, 2002 10:51 AM
To: [EMAIL PROTECTED]
Subject: Regular Expression
I have seen a couple of good regular expression questions asked and
answered. Now I have a question. This is probably easy but i'm stuck on it.
All I want to do is take.
my @numbers =
(01,02,03,03,05,08,09,12,14,13,11,18,17,12,15,16,15,16,12,13,14,16,17,22,23,
24,25,25,23,22,21,20);
so on, could be up to 99. Read them in and get a count of all the numbers.
How many 01 or 17 or 22 there are in the array. I saw an answer for doing
this with single numbers such as 2897, but how with double numbers. The out
put i'm looking for is similar.
one - however many times.
two - however many times.
...
...
twenty - however many times.
...
...
You get the picture :-)
Any help would be great.
Thanks in advance. Steve.
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RE: Regular Expression
At 05:59 PM 05/05/2002, you wrote:
Hi Steve,
This may not be the most eficient way, but it seems to work.
Someone will undoubtedly offer another (probably better) way to do it.
This is good, but more than necessary...
Toby
snip
my @numbers = qw(01 02 03 03 05 08 09 12 14 13 11 18 17 12 15 16 15 16 12 13
14 16 17 22 23 24 25 25 23 22 21 20);
my %counts;
for my $num ( @numbers ) {
$counts{$num}++;
}
for my $num ( sort keys %counts ) {
print There were $counts{$num} occurences of $num\n;
}
Notably, no need to check for the existence of the hash element before incrementing
it, even with strict.
occur_in_array(\@numbers);
sub occur_in_array
{
my $aRef = shift;
foreach (@$aRef)
{
if (!exists($counts{$_}))
{
$counts{$_} = 1;
}
else
{
$counts{$_}++;
}
}
}
while (($key,$value) = each %counts)
{
print $key = $value\n;
}
/snip
-Original Message-
From: steve silvers [mailto:[EMAIL PROTECTED]]
Sent: Monday, May 06, 2002 10:51 AM
To: [EMAIL PROTECTED]
Subject: Regular Expression
I have seen a couple of good regular expression questions asked and
answered. Now I have a question. This is probably easy but i'm stuck on it.
All I want to do is take.
my @numbers =
(01,02,03,03,05,08,09,12,14,13,11,18,17,12,15,16,15,16,12,13,14,16,17,22,23,
24,25,25,23,22,21,20);
so on, could be up to 99. Read them in and get a count of all the numbers.
How many 01 or 17 or 22 there are in the array. I saw an answer for doing
this with single numbers such as 2897, but how with double numbers. The out
put i'm looking for is similar.
one - however many times.
two - however many times.
...
...
twenty - however many times.
...
...
You get the picture :-)
Any help would be great.
Thanks in advance. Steve.
_
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http://photos.msn.com/support/worldwide.aspx
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Re: Regular Expression Question
Try this
#
my $text = this is a website: www.hello-world.com and an e-mail:
[EMAIL PROTECTED];
@found = $text =~ m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g;
print Found something interesting:\n, join \n, @found if @found;
#
which uses the m//g operation in a list context. Explanation from the
perlop manpage:
The /g modifier specifies global pattern matching--that is, matching as many
times as possible within the string. How it behaves depends on the
context. In list context, it returns a list of the substrings matched by
any capturing parentheses in the regular expression. If there are no
parentheses, it returns a list of all the matched strings, as if there
were parentheses around the whole pattern.
In scalar context, each execution of m//g finds the next match, returning true if it
matches, and false if there is
no further match. The position after the last match can be read or set
using the pos() function; see pos in the perlfunc manpage. A failed match normally
resets the search position to the beginning of
the string, but you can avoid that by adding the /c modifier (e.g. m//gc). Modifying
the target string also resets the search position.
Tim
_
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voice (512) 874-5342
fax (512) 874-8500
Joseph Youngquist [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
03/04/2002 02:23 PM
To: [EMAIL PROTECTED]
cc:
Subject:Regular Expression Question
Hello all, I hope some RegEx guru could answer why this stops once it
finds
the first occurrence of the expression.
m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g
This is trying to find web addresses and e-mail addresses from any thing
that is sent to it...
So far, I'm able to get somesite.com, www.somesite.somedomain or
john.doe@somedomain, etc...
The thing that's not working, is of you have more than one of these on a
line sent to it.
so:
my $text = this is a website: www.hello-world.com and an e-mail:
[EMAIL PROTECTED]
if($text =~ m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g) {
print Found something interesting!\n\t$1\n;
}
Thank you for the extra eye-balls,
Joe Y.
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RE: Regular Expression Question
Only problem with using html::parser is I'm not parsing an html document...
There are no html tags to hook onto ... if there were, I'd be happy.
What the domain my script must do is this...say we have some text:
Once upon a time, in a galaxy far, far away. Yogi ([EMAIL PROTECTED]) found a
family where the Chi (for more information see: www.yedi-chi.ye) was strong.
Now, I need to wrap the e-mail addresses and web URLs with the html mark-up
to allow people to click on the link instead of copy and paste
Now if html::parser can re-wrap html on text that'd be slick but I'm not
familiar with the package. Anyone out there know if this is possible with
the package?
Now...if this were a GUI where one was trapping events from a Text control
and adding a style to make the email or URL be blue underlined text, there
wouldn't be a need for html.
Thanks for the idea, I'll poke about with it...if no one sends a yes/no to
the question above.
Joe Y.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of
Jeffrey
Sent: Monday, March 04, 2002 3:51 PM
To: [EMAIL PROTECTED]
Subject: RE: Regular Expression Question
I've never used the module, but you might want to look
at HTML::Parser, and avoid the reinvention of that
round thingy (aka the Wheel). ;)
--- Morse, Richard E. [EMAIL PROTECTED] wrote:
I think that you may need to do this in a while
loop:
my @items
while($text =~ m/your string here/g) {
push @items, $1;
}
print join(\n, @items);
HTH,
Ricky
-Original Message-
From: Joseph Youngquist
[mailto:[EMAIL PROTECTED]]
Sent: Monday 04 March 2002 3:23 PM
To: [EMAIL PROTECTED]
Subject: Regular Expression Question
Hello all, I hope some RegEx guru could answer why
this stops once it finds
the first occurrence of the expression.
m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g
This is trying to find web addresses and e-mail
addresses from any thing
that is sent to it...
So far, I'm able to get somesite.com,
www.somesite.somedomain or
john.doe@somedomain, etc...
The thing that's not working, is of you have more
than one of these on a
line sent to it.
so:
my $text = this is a website: www.hello-world.com
and an e-mail:
[EMAIL PROTECTED]
if($text =~
m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g)
{
print Found something interesting!\n\t$1\n;
}
Thank you for the extra eye-balls,
Joe Y.
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=
Jeffrey Hottle
nkuvu at monkey yahoo dot com
(remove the animal to email me!)
__
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Yahoo! Sports - sign up for Fantasy Baseball
http://sports.yahoo.com
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RE: Regular Expression Question
Thanks, this did the trick...although when I tried this before I got an
infinit loop...bahh just my programing :)...any how, the
while($text =~ m/blah/g) {
...
}
works.
Again, thank you
Joe Y.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of
Morse, Richard E.
Sent: Monday, March 04, 2002 3:32 PM
To: 'Joseph Youngquist'; [EMAIL PROTECTED]
Subject: RE: Regular Expression Question
I think that you may need to do this in a while loop:
my @items
while($text =~ m/your string here/g) {
push @items, $1;
}
print join(\n, @items);
HTH,
Ricky
-Original Message-
From: Joseph Youngquist [mailto:[EMAIL PROTECTED]]
Sent: Monday 04 March 2002 3:23 PM
To: [EMAIL PROTECTED]
Subject: Regular Expression Question
Hello all, I hope some RegEx guru could answer why this stops once it finds
the first occurrence of the expression.
m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g
This is trying to find web addresses and e-mail addresses from any thing
that is sent to it...
So far, I'm able to get somesite.com, www.somesite.somedomain or
john.doe@somedomain, etc...
The thing that's not working, is of you have more than one of these on a
line sent to it.
so:
my $text = this is a website: www.hello-world.com and an e-mail:
[EMAIL PROTECTED]
if($text =~ m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g) {
print Found something interesting!\n\t$1\n;
}
Thank you for the extra eye-balls,
Joe Y.
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Re: Regular Expression Question
Resending because I never saw the message post. Sorry if duplicate.
Try this
#
my $text = this is a website: www.hello-world.com and an e-mail:
[EMAIL PROTECTED];
@found = $text =~ m/\s+((?:[\w\d\-\~]{2,}[@|\.](?:[\w\d\-\~]{2,}\.?)+))/g;
print Found something interesting:\n, join \n, @found if @found;
#
which uses the m//g operation in a list context. Explanation from the
perlop manpage:
The /g modifier specifies global pattern matching--that is, matching as many
times as possible within the string. How it behaves depends on the
context. In list context, it returns a list of the substrings matched by
any capturing parentheses in the regular expression. If there are no
parentheses, it returns a list of all the matched strings, as if there
were parentheses around the whole pattern.
In scalar context, each execution of m//g finds the next match, returning true if it
matches, and false if there is
no further match. The position after the last match can be read or set
using the pos() function; see pos in the perlfunc manpage. A failed match normally
resets the search position to the beginning of
the string, but you can avoid that by adding the /c modifier (e.g. m//gc). Modifying
the target string also resets the search position.
Tim
_
Tim Moose | T R I L O G Y
voice (512) 874-5342
fax (512) 874-8500
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RE: regular expression
Bill,
You misunderstood me, I need to convert
$name[0] = tom???.???;
$name[1] = tom*.???;
into
$name[0] = tom[a-z|A-Z]{3}.[a-z|A-Z]{2};
$name[1] = tom*.[0-9]{3};
So I can build a grep command to get all the files in this directory.
the purpose is to look through all files and check it's attributes.
PC
-Original Message-
From: $Bill Luebkert [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 31, 2002 3:23 PM
To: Wang, Pin-Chieh
Cc: [EMAIL PROTECTED]
Subject: Re: regular expression
Wang, Pin-Chieh wrote:
Hi,
Is there a simple way to convert a variable from wild card represention
into
regular expression?
i.e.
$name[0] = tom???.???;
$name[1] = tom*.???;
into
$name[0] = tom[a-z|A-Z]{3}.[a-z|A-Z]{2};
$name[1] = tom*.[0-9]{3};
Actually that's not correct. The '?' can be converted to '.' and
the '*' can be converted to '.*' :
foreach (@name) {
s/\?/\./g;
s/\*/\.\*/g;
}
Depending on OS, certain characters aren't legal for the '.'. So
you could change the '.' to a character class that represents the legal
characters that a filename can take on (or a negation of the ones it
can't). But that shouldn't be necessary since the filesystem won't
have any files in it that aren't legal and the '.' should work fine.
--
,-/- __ _ _ $Bill Luebkert ICQ=14439852
(_/ / )// // DBE Collectibles Mailto:[EMAIL PROTECTED]
/ ) /-- o // // http://dbecoll.tripod.com/ (Free site for Perl)
-/-' /___/__/_/_ Castle of Medieval Myth Magic
http://www.todbe.com/
**
This email and any files transmitted with it from the ElPaso
Corporation are confidential and intended solely for the
use of the individual or entity to whom they are addressed.
If you have received this email in error please notify the
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RE: regular expression
$name[0] = tom???.???;
$name[1] = tom*.???;
into
$name[0] = tom[a-z|A-Z]{3}.[a-z|A-Z]{2};
$name[1] = tom*.[0-9]{3};
So I can build a grep command to get all the files in this directory.
the purpose is to look through all files and check it's attributes.
I believe what you're needing is a regular expression to convert your
wildcard into a regular expression.
Try this for a start:
$name[0] =~ s|\.|\\.|g;
$name[0] =~ s|\?|.|g;
$name[0] =~ s|\*|.*|g;
I realize this doesn't even come close to being complete, but it's a start.
jpt
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RE: Regular expression for an IPv4 address
The following:
use strict;
sub try1;
try1 001.001.001.001;
try1 1.1.101.1;
try1 002.2.102.001;
exit(0);
sub try1{
my $str = shift;
my $str2 = $str;
$str2 =~ s/(^|\.)(0*)(\d*)/$1$3/go;
print in: '$str' out: '$str2'\n;
}
gives:
C:\TEMPc:\temp\t1.pl
in: '001.001.001.001' out: '1.1.1.1'
in: '1.1.101.1' out: '1.1.101.1'
in: '002.2.102.001' out: '2.2.102.1'
-Original Message-
From: CARPENTER,STEPHEN (HP-Corvallis,ex1)
[mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 16, 2002 7:25 PM
To: '[EMAIL PROTECTED]'
Subject: Regular expression for an IPv4 address
Hi,
I'm a beginner with Perl and I'm sure this has been
done before: I
would like to take a string like this '001.202.033.400' and
turn it into
'1.202.33.400'. In other words, strip leading zeros from each
octet of an IP
address. I could just split each piece into a different
variable, do it, and
then recombine, but is seems like a one line regular
expression could work
here instead. Thanks to all who reply!
StephenC.
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RE: Regular expression for an IPv4 address
Try for example
$ip = join '.',map {int} split /\./,$oldip;
or (if you resist on a regular expression)
$oldip =~ s/(?:^0+|(\.)0+([1-9]+)|(\.0)00)/$1.$2 || $3/eg;
JB
-Original Message-
From: CARPENTER,STEPHEN (HP-Corvallis,ex1)
[mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 17, 2002 1:25 AM
To: '[EMAIL PROTECTED]'
Subject: Regular expression for an IPv4 address
Hi,
I'm a beginner with Perl and I'm sure this has been done before:
I
would like to take a string like this '001.202.033.400' and turn it into
'1.202.33.400'. In other words, strip leading zeros from each octet of
an IP
address. I could just split each piece into a different variable, do it,
and
then recombine, but is seems like a one line regular expression could
work
here instead. Thanks to all who reply!
StephenC.
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RE: Regular expression for an IPv4 address
I am sure one of the experts can make these two regex lines into one, but here it is.. $z = '001.022.003.040'; print $z\n; $z =~ s/^0+//; $z =~ s/\.0+/\./g; print $z\n; ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] http://listserv.ActiveState.com/mailman/listinfo/perl-win32-users
Re: Regular expression help
On Sat, 1 Dec 2001, linkagent wrote:
- Original Message -
From: $Bill Luebkert [EMAIL PROTECTED]
To: linkagent [EMAIL PROTECTED]
linkagent wrote:
I need members help on this;
Q1)As far as I know, \d* means match either 0 or more digits, since
/(\d*)/ match 1006326869812 , therefore
I could not see how /(\d*)(\d{4})(\d{5})/) could seperate 1006326869812
into
(1006)(3268)(69812)
The last two parts of the regex pick up 4 and 5 digits resp. The first
picks up all the rest since it's 0 or more. Obviously some back-tracking
has to occur since the first length is unknown. It might be faster to
anchor the back end:
/(\d*)(\d{4})(\d{5})$/
Correct me if I am wrong;
Therefore am I right to say that the matching sequence starts from the back
first (which is not what I read from the books about matching / /).
i.e in the following match /(\d*)(\d{4})(\d{5})$/
the regexes look for $ first;
then followed by (\d{5}) ;
then followed by (\d{4})
then (\d*)
Q2)Out of curiosity, I tried the undermentioned but I could not
understand why on the 3rd iteration it print only 3 instead of
35968322963568389 :-
for $number (1006326869812, 563296853235993 , 35968322963568389){
print $1\n if ($number =~ /(\d*)/);
}
The third number is too large to express as an integer.
Try putting quotes around it so it's treated as a string.
I thought if the number is too large, it will display something like this
3e0whatever-number, nevertheless, I will read on this.
Which may explain why the last number matched only on the leading 3 digit.
To find out for sure, the print statement could be changed to:
print $1 number=$number\n if $number=~/(\d*)/;
Actually on my PC the third number printed out as: 3.59683229635684e+016
Note that the character in the second-most significant position of the
mantissa is the . character which doesn't match the \d* regex pattern.
Bill's suggestion to surround the numbers with quotes will allow the
entire string of digits to match.
[EMAIL PROTECTED] Carl Jolley
All opinions are my own and not necessarily those of my employer
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Re: Regular expression help
- Original Message -
From: $Bill Luebkert [EMAIL PROTECTED]
To: linkagent [EMAIL PROTECTED]
linkagent wrote:
I need members help on this;
Q1)As far as I know, \d* means match either 0 or more digits, since
/(\d*)/ match 1006326869812 , therefore
I could not see how /(\d*)(\d{4})(\d{5})/) could seperate 1006326869812
into
(1006)(3268)(69812)
The last two parts of the regex pick up 4 and 5 digits resp. The first
picks up all the rest since it's 0 or more. Obviously some back-tracking
has to occur since the first length is unknown. It might be faster to
anchor the back end:
/(\d*)(\d{4})(\d{5})$/
Correct me if I am wrong;
Therefore am I right to say that the matching sequence starts from the back
first (which is not what I read from the books about matching / /).
i.e in the following match /(\d*)(\d{4})(\d{5})$/
the regexes look for $ first;
then followed by (\d{5}) ;
then followed by (\d{4})
then (\d*)
Q2)Out of curiosity, I tried the undermentioned but I could not
understand why on the 3rd iteration it print only 3 instead of
35968322963568389 :-
for $number (1006326869812, 563296853235993 , 35968322963568389){
print $1\n if ($number =~ /(\d*)/);
}
The third number is too large to express as an integer.
Try putting quotes around it so it's treated as a string.
I thought if the number is too large, it will display something like this
3e0whatever-number, nevertheless, I will read on this.
Thank You.
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Re: Regular expression help
On Sat, 1 Dec 2001, linkagent wrote:
- Original Message -
From: $Bill Luebkert [EMAIL PROTECTED]
To: linkagent [EMAIL PROTECTED]
linkagent wrote:
I need members help on this;
Q1)As far as I know, \d* means match either 0 or more digits, since
/(\d*)/ match 1006326869812 , therefore
I could not see how /(\d*)(\d{4})(\d{5})/) could seperate 1006326869812
into
(1006)(3268)(69812)
The last two parts of the regex pick up 4 and 5 digits resp. The first
picks up all the rest since it's 0 or more. Obviously some back-tracking
has to occur since the first length is unknown. It might be faster to
anchor the back end:
/(\d*)(\d{4})(\d{5})$/
Correct me if I am wrong;
Therefore am I right to say that the matching sequence starts from the back
first (which is not what I read from the books about matching / /).
Let me see if I understand it, because this had me scrathing my head
for a while, too (I'm not so great at regex). The regex
/(\d*)(\d{4})(\d{5})/
Means match any number of digits that is followed by a string of 4
digits, that is followed by a string of 5 digits, (and slop those three
things into $1, $2, and $3).
So, using, for example, 1006326869812, 1006 is the only string of digits
that is followed by a string of 4 digits (3268), that is followed by a
string of 5 digits (69812).
Now, the regex /(\d*)(\d{4})(\d{5})/ means almost the same thing: match
any number of digits that is followed by a string of 4 digits, that is
followed by a string of 5 digits that ends at the end of the string (and
slop into $1, $2 and $3).
Same thing in your context, but by telling perl up front about that
end-of-string, perl should start its processing using that part of the
regex, rather than spend a lot of computation trying all the possible
permutations that might match that initial (\d*).
Q2)Out of curiosity, I tried the undermentioned but I could not
understand why on the 3rd iteration it print only 3 instead of
35968322963568389 :-
The third number is too large to express as an integer.
Try putting quotes around it so it's treated as a string.
I thought if the number is too large, it will display something like this
3e0whatever-number, nevertheless, I will read on this.
That would be consistent with matching only the first (\d*), and matching
it only to '3'.
--
Terry Carroll| There ain't no such thing as a free lunch.
Santa Clara, CA | - Washington Legal Foundation v. Legal
[EMAIL PROTECTED] | Foundation of Washington, no. 98-35154
Modell delendus est | (9th Cir. Nov. 14, 2001) (Kozinski, dissenting)
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Re: Regular expression help
linkagent [EMAIL PROTECTED] wrote: Correct me if I am wrong; Therefore am I right to say that the matching sequence starts from the back first (which is not what I read from the books about matching / /). No, the matcher starts from the front, but when it fails it backtracks to see if there's a match with earlier * and + being less greedy. I thought if the number is too large, it will display something like this 3e0whatever-number, nevertheless, I will read on this. It does display something like that. Try print 35968322963568389. You'll get something like 3.59683229635684e+016, which is what Perl does when asked to convert the number 35968322963568389 to a string. Regex matching works on strings, so the '\d*' stops matching at the '.'. -- Keith C. Ivey [EMAIL PROTECTED] Washington, DC ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] http://listserv.ActiveState.com/mailman/listinfo/perl-win32-users
Re: Regular expression help
linkagent wrote:
Correct me if I am wrong;
Therefore am I right to say that the matching sequence starts from the back
first (which is not what I read from the books about matching / /).
i.e in the following match /(\d*)(\d{4})(\d{5})$/
the regexes look for $ first;
then followed by (\d{5}) ;
then followed by (\d{4})
then (\d*)
To the best of my knowledge, the engine always starts at the front.
Anchoring at the back makes the back-tracking easier/simpler.
--
,-/- __ _ _ $Bill Luebkert ICQ=14439852
(_/ / )// // DBE Collectibles Mailto:[EMAIL PROTECTED]
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Re: Regular expression help
- Original Message -
From: Ron Hartikka [EMAIL PROTECTED]
for $number (1006326869812, 563296853235993 , 35968322963568389){
print $1-$2-$3\n if ($number =~ /(\d*)(\d{4})(\d{5})/);
I need members help on this;
Q1)As far as I know, \d* means match either 0 or more digits, since
/(\d*)/ match 1006326869812 , therefore
I could not see how /(\d*)(\d{4})(\d{5})/) could seperate 1006326869812 into
(1006)(3268)(69812)
Q2)Out of curiosity, I tried the undermentioned but I could not
understand why on the 3rd iteration it print only 3 instead of
35968322963568389 :-
for $number (1006326869812, 563296853235993 , 35968322963568389){
print $1\n if ($number =~ /(\d*)/);
}
Results :-
1006326869812
563296853235993
3
Thanks
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Re: Regular expression help
linkagent wrote:
- Original Message -
From: Ron Hartikka [EMAIL PROTECTED]
for $number (1006326869812, 563296853235993 , 35968322963568389){
print $1-$2-$3\n if ($number =~ /(\d*)(\d{4})(\d{5})/);
I need members help on this;
Q1)As far as I know, \d* means match either 0 or more digits, since
/(\d*)/ match 1006326869812 , therefore
I could not see how /(\d*)(\d{4})(\d{5})/) could seperate 1006326869812 into
(1006)(3268)(69812)
The last two parts of the regex pick up 4 and 5 digits resp. The first
picks up all the rest since it's 0 or more. Obviously some back-tracking
has to occur since the first length is unknown. It might be faster to
anchor the back end:
/(\d*)(\d{4})(\d{5})$/
Q2)Out of curiosity, I tried the undermentioned but I could not
understand why on the 3rd iteration it print only 3 instead of
35968322963568389 :-
for $number (1006326869812, 563296853235993 , 35968322963568389){
print $1\n if ($number =~ /(\d*)/);
}
The third number is too large to express as an integer.
Try putting quotes around it so it's treated as a string.
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RE: Regular expression help
for $number (1006326869812, 563296853235993 , 35968322963568389){
print $1-$2-$3\n if ($number =~ /(\d*)(\d{4})(\d{5})/);
}
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of
[EMAIL PROTECTED]
Sent: Thursday, November 29, 2001 3:32 PM
To: [EMAIL PROTECTED]
Subject: Regular expression help
I need help creating a regular expression to do the following.
I have the following numbers:
1006326869812
563296853235993
35968322963568389
and it needs to be broken up like this
1006-3268-69812
563296-8532-35993
35968322-9635-68389
Notice the second group of numbers is always 4 places and the
last group is
always 5 places.
I'm stumped and not to good with RE's. Any pointers would be great.
Rob
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RE: Regular expression help
[EMAIL PROTECTED] wrote:
I need help creating a regular expression to do the following.
I have the following numbers:
1006326869812
563296853235993
35968322963568389
and it needs to be broken up like this
1006-3268-69812
563296-8532-35993
35968322-9635-68389
Notice the second group of numbers is always 4 places and the
last group is
always 5 places.
I'm stumped and not to good with RE's. Any pointers would be great.
s/(\d+)(\d{4})(\d{5})/$1-$2-$3/;
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RE: Regular expression help
You don't even need a regex although you could use one...
# untested
my $num = 92739874598745;
$num =~ /^(\d*)(d{4})(\d{5})$/;
my ($n1, $n2, $n3) = ($1, $2, $3);
Or you could do this...
# untested
my $num = 92739874598745;
my $n1 = substr($num, 0, length($num) - 9);
my $n2 = substr($num, -9, 4);
my $n3 = substr($num, -5);
Also add some error checking unless you are sure that the number is at least
9 chars long.
Rob
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Thursday, November 29, 2001 3:32 PM
To: [EMAIL PROTECTED]
Subject: Regular expression help
I need help creating a regular expression to do the following.
I have the following numbers:
1006326869812
563296853235993
35968322963568389
and it needs to be broken up like this
1006-3268-69812
563296-8532-35993
35968322-9635-68389
Notice the second group of numbers is always 4 places and the last group is
always 5 places.
I'm stumped and not to good with RE's. Any pointers would be great.
Rob
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Re: Regular expression help
How about something like
s/(\d+)(\d{4})(\d{5})/$1-$2-$3/
?
--- [EMAIL PROTECTED] wrote:
I need help creating a regular expression to do the
following.
I have the following numbers:
1006326869812
563296853235993
35968322963568389
and it needs to be broken up like this
1006-3268-69812
563296-8532-35993
35968322-9635-68389
Notice the second group of numbers is always 4
places and the last group is
always 5 places.
I'm stumped and not to good with RE's. Any pointers
would be great.
Rob
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=
Jeffrey Hottle
nkuvu at yahoo dot com
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Re: Regular expression help
[EMAIL PROTECTED] wrote:
I need help creating a regular expression to do the following.
I have the following numbers:
1006326869812
563296853235993
35968322963568389
and it needs to be broken up like this
1006-3268-69812
563296-8532-35993
35968322-9635-68389
Notice the second group of numbers is always 4 places and the last group
is
always 5 places.
I'm stumped and not to good with RE's. Any pointers would be great.
try this:
$number=~s/^(\d+)(\d{4})(\d{5})$/$1-$2-$3/;
hth,
-dave
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RE: Regular expression help
Thanks to all those who responded Rob -Original Message- From: Plane, Robert Sent: Thursday, November 29, 2001 3:32 PM To: '[EMAIL PROTECTED]' Subject: Regular expression help I need help creating a regular expression to do the following. I have the following numbers: 1006326869812 563296853235993 35968322963568389 and it needs to be broken up like this 1006-3268-69812 563296-8532-35993 35968322-9635-68389 Notice the second group of numbers is always 4 places and the last group is always 5 places. I'm stumped and not to good with RE's. Any pointers would be great. Rob ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] http://listserv.ActiveState.com/mailman/listinfo/perl-win32-users
[PBML] Re: Regular expression needed
Hi, I have a file with several lines which begins with:
/bootp/linux/
I would like to match theses lines
I tried this while (INFILE) {
chomp();
/^\/bootp/linux/\/
but it doesn't work how can I do ?
Thanks.
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RE: [PBML] Re: Regular expression needed
Title: RE: [PBML] Re: Regular expression needed
you need to escape all of the /'s in your path with in
the match
/^\/bootp\/linux\//;
should match;
a good technique for building regular expressions is to
match a small portion of what you're looking for. then
after you get that part working continue adding to your
regular expression, if that doesn't match you know something
you just added was faulty.
-Original Message-
From: Jorge Goncalvez [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, October 10, 2001 9:41 AM
To: [EMAIL PROTECTED]
Subject: [PBML] Re: Regular expression needed
Hi, I have a file with several lines which begins with:
/bootp/linux/
I would like to match theses lines
I tried this while (INFILE) {
chomp();
/^\/bootp/linux/\/
but it doesn't work how can I do ?
Thanks.
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RE: [PBML] Re: Regular expression needed
Another method is to specify different delimiters for the match. This avoids LTS (leaning toothpick syndrome): m#^/bootp/linux/#; m%^/bootp/linux/%; m(^/bootp/linux/); Each of the lines above are equivalent -- they're also equivalent to /^\/bootp\/linux\//; --- Vroom, Tim [EMAIL PROTECTED] wrote: you need to escape all of the /'s in your path with in the match /^\/bootp\/linux\//; = Jeffrey [EMAIL PROTECTED] __ Do You Yahoo!? Make a great connection at Yahoo! Personals. http://personals.yahoo.com ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] http://listserv.ActiveState.com/mailman/listinfo/perl-win32-users
RE: [PBML] Re: Regular expression needed
Jeffrey [mailto:[EMAIL PROTECTED]] wrote:
Another method is to specify different delimiters for
the match. This avoids LTS (leaning toothpick
syndrome):
m#^/bootp/linux/#;
m%^/bootp/linux/%;
m(^/bootp/linux/);
Each of the lines above are equivalent -- they're also
equivalent to /^\/bootp\/linux\//;
No one mentioned this: /^\Q/bootp/linux/\E/;
also equivalent, but has an advantage when doing things like this:
$skipdir = '/bootp/linux/';
while (INFILE) {
next if /^\Q$skipdir\E/;
...
}
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