regular expression, this but not that
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. -Jeremy ___ 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 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
Regular expression question
Title: 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 ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Regular expression question
Title: 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 ___ 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
RES: Regular expression question
Title: Regular expression question
hi
wagner,
try
this:
#/perl
-w
$file =
'c:\temp\zips\foo\foo2\foo3\ok.txt'; @pa = split (/\\/, $file);$file =
$pa[$#pa];
for $i
(0..($#pa-1)) {$dir .=
"$pa[$i]\\";}
print "directory:
$dir\n";print "file: $file\n";
fabrício s.
martins
-Mensagem original-De:
[EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]Em nome de
Wagner, David --- Senior Programmer Analyst --- WGOEnviada em:
terça-feira, 1 de agosto de 2006 16:37Para: Cai, Lucy (L.);
perl-win32-users@listserv.ActiveState.com;
perl-unix-users@listserv.ActiveState.com;
[EMAIL PROTECTED];
[EMAIL PROTECTED]Assunto: RE:
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.
**
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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 $Output =~ m/ucmvob/; # returns no. of matches i.e. 1 or 0 -- REMEMBER THE WORLD TRADE CENTER ---= WTC 911 =-- ...ne cede malis 0100 ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Regular expression question
Hi, all,
I have a question about regular expression:
my code is like this:
**
sub IsPVob
{
my ($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
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
Regular Expression
Hi, I'm trying to figure out how to parse a string with special characters. $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. ___ 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
A regular expression question -- Emergency
Hi All,
I am meeting a problem on a regular expression.
My code is like:
$path_elem = $ENV{'CLEARCASE_PN'};
The return value of $path_elem is either
$path_elem = /ccstore/test/test.pl
Or $path_elem = /ccstore/test 1/test 1.pl # there is a space in the
path name including the file name too
There is no rule whether there is a space in the path name or not. How
can I handle it in regular expression, because when I pass the value
$path_elem to a function, if there is a space in the path name, it will
treat it 2 file path name, instead of one.
I used the following code
$path_elem = \$ENV{'CLEARCASE_PN'}\;
It works fine for windows, but does not works fine on unix.
I really appriciate your great help.
Thanks
Lucy
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Re: A regular expression question -- Emergency
On UNIX you have to escape the whitespace, e.g.
/this path/has spaces
would become
/this\ path/has\ spaces
The expression
$path_elem =~ s{(\s)}{\\$1}gio;
should do the trick.
Note that I used {} delimiters (i.e. s{}{}) to make it clearer, since
the data itself contains /s and someone might get confused that a / was
part of the match or substitution, rather than a delimiter. But you
still could use s///, if you prefer. Also, \\ is necessary in the
substituion value, since \ itself is a metacharacter in a regex.
Cheers,
Richard
Cai, Lucy (L.) wrote:
Hi All,
I am meeting a problem on a regular expression.
My code is like:
$path_elem = $ENV{'CLEARCASE_PN'};
The return value of $path_elem is either
$path_elem = /ccstore/test/test.pl
Or $path_elem = /ccstore/test 1/test 1.pl # there is a space in the
path name including the file name too
There is no rule whether there is a space in the path name or not. How
can I handle it in regular expression, because when I pass the value
$path_elem to a function, if there is a space in the path name, it will
treat it 2 file path name, instead of one.
I used the following code
$path_elem = \$ENV{'CLEARCASE_PN'}\;
It works fine for windows, but does not works fine on unix.
I really appriciate your great help.
Thanks
Lucy
-
*** Richard A. Wells, [EMAIL PROTECTED], +1.978.371.7425
*** Reality And Wonder, http://www.raw.com/
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Re: A regular expression question -- Emergency
On UNIX you have to escape the whitespace, e.g.
/this path/has spaces
would become
/this\ path/has\ spaces
The expression
$path_elem =~ s{(\s)}{\\$1}gio;
should do the trick.
Note that I used {} delimiters (i.e. s{}{}) to make it clearer, since
the data itself contains /s and someone might get confused that a / was
part of the match or substitution, rather than a delimiter. But you
still could use s///, if you prefer. Also, \\ is necessary in the
substituion value, since \ itself is a metacharacter in a regex.
Cheers,
Richard
Cai, Lucy (L.) wrote:
Hi All,
I am meeting a problem on a regular expression.
My code is like:
$path_elem = $ENV{'CLEARCASE_PN'};
The return value of $path_elem is either
$path_elem = /ccstore/test/test.pl
Or $path_elem = /ccstore/test 1/test 1.pl # there is a space in the
path name including the file name too
There is no rule whether there is a space in the path name or not. How
can I handle it in regular expression, because when I pass the value
$path_elem to a function, if there is a space in the path name, it will
treat it 2 file path name, instead of one.
I used the following code
$path_elem = \$ENV{'CLEARCASE_PN'}\;
It works fine for windows, but does not works fine on unix.
I really appriciate your great help.
Thanks
Lucy
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Re: A regular expression question -- Emergency
Richard A. Wells wrote:
On UNIX you have to escape the whitespace, e.g.
/this path/has spaces
would become
/this\ path/has\ spaces
The expression
$path_elem =~ s{(\s)}{\\$1}gio;
The /io is not needed/wanted.
should do the trick.
Note that I used {} delimiters (i.e. s{}{}) to make it clearer, since
the data itself contains /s and someone might get confused that a / was
part of the match or substitution, rather than a delimiter. But you
still could use s///, if you prefer. Also, \\ is necessary in the
substituion value, since \ itself is a metacharacter in a regex.
I think it's easier to read this:
my $win32 = $^O =~ /Win32/i;
my $path_elem = $ENV{CLEARCASE_PN};
$path_elem =~ s/ /\ /g if not $win32;
PS: Top-posting is something we try to avoid in these lists.
Hi All,
I am meeting a problem on a regular expression.
My code is like:
$path_elem = $ENV{'CLEARCASE_PN'};
The return value of $path_elem is either
$path_elem = /ccstore/test/test.pl
Or $path_elem = /ccstore/test 1/test 1.pl # there is a space in the
path name including the file name too
There is no rule whether there is a space in the path name or not. How
can I handle it in regular expression, because when I pass the value
$path_elem to a function, if there is a space in the path name, it will
treat it 2 file path name, instead of one.
I used the following code
$path_elem = \$ENV{'CLEARCASE_PN'}\;
It works fine for windows, but does not works fine on unix.
I really appriciate your great help.
--
,-/- __ _ _ $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)
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Re: A regular expression question -- Emergency
Richard A. Wells wrote:
$Bill Luebkert wrote:
Richard A. Wells wrote:
[...]
The expression
$path_elem =~ s{(\s)}{\\$1}gio;
The /io is not needed/wanted.
Right about the /i. That's conventional laziness on my part, as it is
usually what I want, though it would be detrimental in this case.
Not detrimental - just confusing and unnecessary since there are no
printable characters in the RE.
And I always use /o unless I'm intentionally interpolating in my
pattern. Perhaps the regex engine is smart enough these days to infer
that from the pattern, but it's another piece of conventional laziness.
I almost never use /o unless I'm trying to eke out the last µs from
some code and then it's meaningless if there is no variable to
interpolate in the RE which again confuses the reader.
I think it's easier to read this:
my $win32 = $^O =~ /Win32/i;
my $path_elem = $ENV{CLEARCASE_PN};
$path_elem =~ s/ /\ /g if not $win32;
Sure. It's a choice. But s/(\s)/\$1/g will handle other whitespace, too.
The problem stmt didn't include other WS and I would hope nobody would be
stupid enough to use any other WS in a filename - I refuse to even use a
space. ;)
--
,-/- __ _ _ $Bill LuebkertMailto:[EMAIL PROTECTED]
(_/ / )// // DBE CollectiblesMailto:[EMAIL PROTECTED]
/ ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/
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Re: A regular expression question -- Emergency
$Bill Luebkert wrote:
Richard A. Wells wrote:
[...]
The _expression_
$path_elem =~ s{(\s)}{\\$1}gio;
The /io is not needed/wanted.
Right about the /i. That's conventional laziness on my part, as it is
usually what I want, though it would be detrimental in this case.
And I always use /o unless I'm intentionally interpolating in my
pattern. Perhaps the regex engine is smart enough these days to infer
that from the pattern, but it's another piece of conventional laziness.
I think it's easier to read this:
my $win32 = $^O =~ /Win32/i;
my $path_elem = $ENV{CLEARCASE_PN};
$path_elem =~ s/ /\ /g if not $win32;
Sure. It's a choice. But s/(\s)/\$1/g will handle other whitespace, too.
PS: Top-posting is something we try to avoid in these lists.
Sorry. Understood.
R.
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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
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
Perl Regular Expression Problem
Perl Regular Expression Problem
We are having a problem with Perl's evaluation of regular expressions.
Here is a code snippet:
# This regular expression is looking for a word ending in s followed
# by something in parentheses
# If the while-block is entered, $1 should contain the word ending in s,
# and the original word is recreated and assigned to the variable $method
print( PRE METHOD [$method]\n );
while ( $method $method =~ /(\S+)s\s*\(\s*(.+)\s*\)\s*$/ )
{
$method = $1 . s;
print( POST METHOD [$method]\n );
$retval = ExecuteLAMethod ($retval, $method);
.
.
.
Here is some data from the print statements:
PRE METHOD [Modules(M6833x LA)]
POST METHOD [Modules]
This looks good, as expected.
After evaluating the regular expression, $1 contains 'Module'.
The program continues:
PRE METHOD [Modules(M6833x LA)]
POST METHOD [M6833x LAs]
Here is the problem.
The $method variable contains 'Modules(M6833x LA)' and after
evaluating the regular expression, $1 contains 'M6833x LA'.
I would have expected $1 to contain 'Module is in the first case.
This anomaly is causing our regression-test suite to fail.
Here is some data regarding when this anomaly occurs:
Perl 5.6.1 Build 635 - error Never occurs
Perl 5.8.3 Build 809 - error Always occurs
Perl 5.8.7 Build 813 - error occurs on one machine, but does Not occur on
another
As you can see, we are constrained to an old version of Perl in order to
keep our test suite running.
What is the best way to troubleshoot this kind of error?
Is this a mis-use of Perl variables?
Is there a way to see what is going on inside Perl?
Someone at ActiveState suggested that our code is out-of-date with some
regexpr changes that were made for version 5.8.
Thanks for your help, Paul Schwotzer
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Re: Perl Regular Expression Problem
It works as expected for me and it's build 811. At this point I would take it onto Bugtraq. Something's not right with ur build C:\WINDOWS\Desktopperl $method = 'Modules(M6833x LA)'; $method =~ /(\S+)s\s*\(\s*(.+)\s*\)\s*$/; $method = $1 . s; print( POST METHOD [$method]\n ); POST METHOD [Modules] C:\WINDOWS\Desktopperl -v This is perl, v5.8.6 built for MSWin32-x86-multi-thread (with 3 registered patches, see perl -V for more detail) Copyright 1987-2004, Larry Wall Binary build 811 provided by ActiveState Corp. http://www.ActiveState.com ActiveState is a division of Sophos. Built Dec 13 2004 09:52:01 At 03:38 PM 8/23/05 -0600, [EMAIL PROTECTED] wrote: Here is the problem. The $method variable contains 'Modules(M6833x LA)' and after evaluating the regular expression, $1 contains 'M6833x LA'. I would have expected $1 to contain 'Module is in the first case. Perl 5.6.1 Build 635 - error Never occurs Perl 5.8.3 Build 809 - error Always occurs Perl 5.8.7 Build 813 - error occurs on one machine, but does Not occur on another -- REMEMBER THE WORLD TRADE CENTER ---= WTC 911 =-- ...ne cede males 0100 ___ Perl-Win32-Users mailing list Perl-Win32-Users@listserv.ActiveState.com To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Perl Regular Expression Problem
[EMAIL PROTECTED] wrote:
Perl Regular Expression Problem
We are having a problem with Perl's evaluation of regular expressions.
Here is a code snippet:
# This regular expression is looking for a word ending in s followed
# by something in parentheses
# If the while-block is entered, $1 should contain the word ending in s,
# and the original word is recreated and assigned to the variable $method
print( PRE METHOD [$method]\n );
while ( $method $method =~ /(\S+)s\s*\(\s*(.+)\s*\)\s*$/ )
{
$method = $1 . s;
print( POST METHOD [$method]\n );
$retval = ExecuteLAMethod ($retval, $method);
.
.
.
Here is some data from the print statements:
PRE METHOD [Modules(M6833x LA)]
POST METHOD [Modules]
This looks good, as expected.
After evaluating the regular expression, $1 contains 'Module'.
The program continues:
What do you mean the program continues ? Can you post a complete snippet
that actually fails ?
PRE METHOD [Modules(M6833x LA)]
POST METHOD [M6833x LAs]
Here is the problem.
The $method variable contains 'Modules(M6833x LA)' and after
evaluating the regular expression, $1 contains 'M6833x LA'.
I would have expected $1 to contain 'Module is in the first case.
This anomaly is causing our regression-test suite to fail.
Here is some data regarding when this anomaly occurs:
Perl 5.6.1 Build 635 - error Never occurs
Perl 5.8.3 Build 809 - error Always occurs
Perl 5.8.7 Build 813 - error occurs on one machine, but does Not occur on
another
As you can see, we are constrained to an old version of Perl in order to
keep our test suite running.
What is the best way to troubleshoot this kind of error?
Is this a mis-use of Perl variables?
Is there a way to see what is going on inside Perl?
Someone at ActiveState suggested that our code is out-of-date with some
regexpr changes that were made for version 5.8.
--
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/ ) /-- o // // Castle of Medieval Myth Magic http://www.todbe.com/
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Regular Expression Help Please
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
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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
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: A regular expression question
$Bill Luebkert graced perl with these words of wisdom: B. I want to get 1.62 from the string, how can I do it? Nobody seems to have answered part B. Actually they did. Sam's for one : $string = sct-1.62-1; print $1\n if ($string =~ /^.+-(.+)-.+$/); You mean $1 captures the part in parentheses? Sometimes, trying to decipher regexes is like trying to read chicken scratches :-) -- Ted fedya at bestweb dot net Barney: Hey, Homer, you're late for English. Homer: Who needs English? I'm never going to England. http://www.snpp.com/episodes/7F12.html ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
A regular expression question
Title: A regular expression question Now I have a string like My $string = sct-1.62-1; I have 2 regular _expression_ question here, A. I want to check whether the format is XXX-XXX-XXX, how can I do it? B. I want to get 1.62 from the string, how can I do it? Thanks a lot in advance! Lixin Lixin Cai Security Compliance Tools(SCT) Location: ITek 2WC135 Phone:313-317-4906 email: [EMAIL PROTECTED] ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: A regular expression question
Title: A regular expression question Your format does not match XXX-XXX-XXX so I'll guess you mean three fields delimited by a dash. here's one way to do it $string = "sct-1.62-1"; print "$1\n" if ($string =~ /^.+\-(.+)\-.+$/) -Original Message-From: Cai, Lixin (L.) [mailto:[EMAIL PROTECTED]Sent: Thursday, November 11, 2004 2:37 PMTo: [EMAIL PROTECTED]Subject: A regular _expression_ question Now I have a string like My $string = "sct-1.62-1"; I have 2 regular _expression_ question here, A. I want to check whether the format is "XXX-XXX-XXX", how can I do it? B. I want to get 1.62 from the string, how can I do it? Thanks a lot in advance! Lixin Lixin Cai Security Compliance Tools(SCT) Location: ITek 2WC135 Phone:313-317-4906 email: [EMAIL PROTECTED] Message transport securityby GatewayDefender2:40:28 PM ET - 11/11/2004 ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: A regular expression question
Title: Message or even. . . $string = "sct-1.62-1";print "$1\n" if ($string =~ /^.+-(.+)-.+$/); (no need to use the backslash escape for the dashes; they're not part of a character class. . . Sam Gardner GTO Application Development Keefe, Bruyette Woods, Inc. 212-887-6753 -Original Message-From: Peter Eisengrein [mailto:[EMAIL PROTECTED] Sent: Thursday, November 11, 2004 2:52 PMTo: 'Cai, Lixin (L.)'; [EMAIL PROTECTED]Subject: RE: A regular _expression_ question Your format does not match XXX-XXX-XXX so I'll guess you mean three fields delimited by a dash. here's one way to do it $string = "sct-1.62-1"; print "$1\n" if ($string =~ /^.+\-(.+)\-.+$/) -Original Message-From: Cai, Lixin (L.) [mailto:[EMAIL PROTECTED]Sent: Thursday, November 11, 2004 2:37 PMTo: [EMAIL PROTECTED]Subject: A regular _expression_ question Now I have a string like My $string = "sct-1.62-1"; I have 2 regular _expression_ question here, A. I want to check whether the format is "XXX-XXX-XXX", how can I do it? B. I want to get 1.62 from the string, how can I do it? Thanks a lot in advance! Lixin Lixin Cai Security Compliance Tools(SCT) Location: ITek 2WC135 Phone:313-317-4906 email: [EMAIL PROTECTED] Message transport securityby GatewayDefender2:40:28 PM ET - 11/11/2004 ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: A regular expression question
Cai, Lixin (L.) graced perl with these words of wisdom: My $string = sct-1.62-1; I have 2 regular expression question here, A. I want to check whether the format is XXX-XXX-XXX, how can I do it? B. I want to get 1.62 from the string, how can I do it? Nobody seems to have answered part B. So, assuming you want the middle part (ie, the section between the two dashes), use something like the following (untested): snip my ($string, @line); $string = sct-1.62-1; @line = split /-/, $string; print $line[1]; /snip -- Ted fedya at bestweb dot net Barney: Hey, Homer, you're late for English. Homer: Who needs English? I'm never going to England. http://www.snpp.com/episodes/7F12.html ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: A regular expression question
Ted Schuerzinger wrote: Cai, Lixin (L.) graced perl with these words of wisdom: My $string = sct-1.62-1; I have 2 regular expression question here, A. I want to check whether the format is XXX-XXX-XXX, how can I do it? This format doesn't actually match the example given. Maybe he meant 'AAA-N.NN-N[NN]' or something similar (where [] means optional). B. I want to get 1.62 from the string, how can I do it? Nobody seems to have answered part B. Actually they did. Sam's for one : $string = sct-1.62-1; print $1\n if ($string =~ /^.+-(.+)-.+$/); So, assuming you want the middle part (ie, the section between the two dashes), use something like the following (untested): snip my ($string, @line); $string = sct-1.62-1; @line = split /-/, $string; print $line[1]; /snip -- ,-/- __ _ _ $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
question of regular expression
I have one line like this:
$command = '$ldapsearch -x -LLL -h "cds2.ford.com" -b "ou=People, o=Ford,c=US" "uid=$login" uid fordUNIXid';
If condition
{
$command = '$ldapsearch -LLL -h "cds2.ford.com" -b "ou=People, o=Ford,c=US" "uid=$login" uid fordUNIXid'
# which -x need to remove
}
Could anyone let me know by using regular _expression_, how can I get rid of "-x"?
Thanks in advance!
Lixin
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RE: question of regular expression
Could anyone let me know by using regular expression, how can I get rid of -x? If that's all you really want, and you know that's the only place it can occur, then its no problem to use just s/-x/''/; (substitute '-x' by '') ed c ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
RE: question of regular expression
i wrote, obviously with my eyes closed: s/-x/''/; which of course should be s/-x//; - sorry! ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
LWP and Regular Expression
I'm using LWP::UserAgent to grab data from internal intranet site pages. I
can grab the page just fine with:
$ua = LWP::UserAgent-new;
$ua-proxy(['http', 'ftp'] = 'http://login:[EMAIL PROTECTED]');
$req = HTTP::Request-new('GET',http://google.com;);
$res = $ua-request($req);
print $res-content if $res-is_success;
I need to grab the page data either in an array or scalar and cut out a
certain part of text. So if I use
$all = $res-content if $res-is_success;
Then $all contains all the source code of the page. How do I grab out data.
As an example say I'm grabbing the Google home page. How would I grab out
the bottom ©2004 Google - Searching 4,285,199,774 web pages text? This is
just an example to get me started.
Thanks in advance
Steve
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Re: LWP and Regular Expression
steve silvers wrote:
I'm using LWP::UserAgent to grab data from internal intranet site pages. I
can grab the page just fine with:
$ua = LWP::UserAgent-new;
$ua-proxy(['http', 'ftp'] = 'http://login:[EMAIL PROTECTED]');
$req = HTTP::Request-new('GET',http://google.com;);
$res = $ua-request($req);
print $res-content if $res-is_success;
I need to grab the page data either in an array or scalar and cut out a
certain part of text. So if I use
$all = $res-content if $res-is_success;
Then $all contains all the source code of the page. How do I grab out data.
As an example say I'm grabbing the Google home page. How would I grab out
the bottom ©2004 Google - Searching 4,285,199,774 web pages text? This is
just an example to get me started.
You could use a module or a regex.
Modules like: HTML::Parse, HTML::Parser, HTML::TreeBuilder, etc.
Regex:
$all =~ /Google - Searching ([\d,]+) web pages/s;
my $numpages = $1;
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RE: Regular expression to test for numeric values
So I get: /^-?(?:\d+\.?\d*|\.\d+)$/ 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
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
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
==
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
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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
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
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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Your favorite stores, helpful shopping tools and great gift ideas.
Experience the convenience of buying online with [EMAIL PROTECTED]
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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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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 ## ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
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: Another regular expression question
At 10:28 AM 3/20/2003 -0500, you wrote:
Filename = Base-02.04.1.20.5.002-xlite_katana_free.ndu;
I want to get extension of the file name which is ndu, but I always get
04.1.20.5.002-xlite_katana_free.ndu instead of ndu.
The sub is like:
sub extension
{
my $path = shift;
my $ext = fileparse($path, '\..*?'))[2];
$ext =~ s/^\.//;
return $ext;
}
what is wrong with it?
I really like to use split for that task
Tobias
my $file=Base-02.04.1.20.5.002-xlite_katana_free.ndu;
print extension for $file is ,extension($file),\n;
sub extension
{
(split(/\./,$_[0]))[-1];
}
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Re: Another regular expression question
Hi!
On 20 Mar 2003 at 10:28, [EMAIL PROTECTED] wrote:
Filename = Base-02.04.1.20.5.002-xlite_katana_free.ndu;
I want to get extension of the file name which is ndu, but I always
get 04.1.20.5.002-xlite_katana_free.ndu instead of ndu.
:
Try the code below. i think it works.
Regards,
Rodrigo
code
use strict;
sub extension
{
my $path = shift;
my @all= split(/\./, $path);
return @all 1?pop @all: undef;
}
my $fName= 'Base-02.04.1.20.5.002-xlite_katana_free.ndu';
print The extension of \$fName\ is \, extension($fName), \\n;
print Type anothe file name: ;
my $otherName= ;
chomp $otherName;
my $ext= extension($otherName);
if ( defined $ext )
{ print \$otherName\ extension is \$ext\\n; }
else
{ print \$otherName\ has no extension!!\n; }
/code
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RE: Another regular expression question
You could use conventional hacking. rindex will get the last '.' and substr will
hand back the remainder.
print extension(Base-02.04.1.20.5.002-xlite_katana_free.ndu);
sub extension
{
return (substr($_[0], rindex($_[0], '.')));
}
-Scott
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Thursday, March 20, 2003 9:28 AM
To: [EMAIL PROTECTED]
Subject: Another regular expression question
Filename = Base-02.04.1.20.5.002-xlite_katana_free.ndu;
I want to get extension of the file name which is ndu, but I always get
04.1.20.5.002-xlite_katana_free.ndu instead of ndu.
The sub is like:
sub extension
{
my $path = shift;
my $ext = fileparse($path, '\..*?'))[2];
$ext =~ s/^\.//;
return $ext;
}
what is wrong with it?
Thanks
Lixin
-Original Message-
From: Randy W. Sims [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 19, 2003 6:24 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: an regular expression question
On 3/19/2003 6:16 PM, [EMAIL PROTECTED] wrote:
All,
A question:
A file like C:\temp\test.txt, how can I get test.txt from this path by
regular expression? I can do it by split function, but I hope I can learn
how to do it by regular expression.
Thanks a lot in advance!
Lixin
use File::Basename;
my $fullpath = 'C:\temp\test.txt';
my $basename = basename( $fullpath );
Randy.
--
A little learning is a dang'rous thing;
Drink deep, or taste not the Pierian spring;
There shallow draughts intoxicate the brain;
And drinking largely sobers us again.
- Alexander Pope
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Re: Another regular expression question
On Thu, 20 Mar 2003 [EMAIL PROTECTED] wrote:
Filename = Base-02.04.1.20.5.002-xlite_katana_free.ndu;
I want to get extension of the file name which is ndu, but I always get
04.1.20.5.002-xlite_katana_free.ndu instead of ndu.
The sub is like:
sub extension
{
my $path = shift;
my $ext = fileparse($path, '\..*?'))[2];
$ext =~ s/^\.//;
return $ext;
}
what is wrong with it?
Thanks
Lixin
-Original Message-
From: Randy W. Sims [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 19, 2003 6:24 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: an regular expression question
On 3/19/2003 6:16 PM, [EMAIL PROTECTED] wrote:
All,
A question:
A file like C:\temp\test.txt, how can I get test.txt from this path by
regular expression? I can do it by split function, but I hope I can learn
how to do it by regular expression.
Thanks a lot in advance!
Lixin
use File::Basename;
my $fullpath = 'C:\temp\test.txt';
my $basename = basename( $fullpath );
Randy.
Your best bet is to use a regex match on the extension anchored at the end
of the string. If your extension is always 3 characters, this would work:
($extension)=$path=~/\.(\w{3})$/;
If the extention can be of arbirtrary length then you could use
($extension)=$path=~/\.(\w+)$/;
[EMAIL PROTECTED] Carl Jolley
All opinions are my own and not necessarily those of my employer
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Beginner Regular Expression Question
I am trying to figure out how to specify or in a regular expression. Basically, I am specifying the Prompt portion of a telnet session using Net::Telnet, and I want to match several possible strings that would make up the prompts. An explanation below: '/br[0-9][0-9]/' or '/mar[0-9][0-9]/' or '/tr[0-9][0-9]/' or '/ber[0-9]/' Does anyone know if this can be done, and if so how? Thanks. Guy H. Lupi ___ Perl-Win32-Users mailing list [EMAIL PROTECTED] To unsubscribe: http://listserv.ActiveState.com/mailman/mysubs
Re: Beginner Regular Expression Question
Net::Telnet, and I want to match several possible strings that would make
up
the prompts. An explanation below:
'/br[0-9][0-9]/' or '/mar[0-9][0-9]/' or '/tr[0-9][0-9]/' or
'/ber[0-9]/'
One way is:
if($string =~ /(?:b|t|be|ma)r[0-9]{2}/) {
...
}
The operative operator (heh) here is '|'
Note, no need to repeat your character-class, tell the RE exactly what you
mean, instead.
For example:
my @strs = ('mar55','br99','tr51','ber00','for11');
foreach my $str (@strs) {
my $ma = didn't;
$ma = did if($str =~ /(?:b|t|be|ma)r[0-9]{2}/);
print(String $str $ma match the RE\n);
}
Prints the following:
C:\source\crapperl ormatch.pl
String mar55 did match the RE
String br99 did match the RE
String tr51 did match the RE
String ber00 did match the RE
String for11 didn't match the RE
!c
C. Church
http://www.digitalKOMA.com/church/
http://www.DroneColony.com/
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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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are addressed. If you have received this email in error please notify
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AW: Regular Expression matching problem
\b is the answer:
--
while (DATA) {
chomp;
if (/\bwilma\b/) {
print wilma was mentioned\n;
}
}
__DATA__
wilma
wimagren was here
twilma was type wilma
wilma
wilma
wilma
twowilmase
--
Output:
wilma was mentioned
wilma was mentioned
wilma was mentioned
wilma was mentioned
wilma was mentioned
regards
Friedel Wittrock
-Ursprungliche Nachricht-
Von: Phil Ritchie [mailto:[EMAIL PROTECTED]
Gesendet: Montag, 10. Marz 2003 09:59
An: Electron One
Cc: [EMAIL PROTECTED]
Betreff: 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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are addressed. If you have received this email in error please notify
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RE: Another Regular expression problem
-Original Message-
From: Electron One [mailto:[EMAIL PROTECTED]
Sent: Monday, March 10, 2003 12:12 PM
To: [EMAIL PROTECTED]
Cc: Electron One
Subject: Another Regular expression problem
Hello Everyone,
I have a perl file that has this,
PerlFile.pl###
#!/usr/bin/perl
while(){
chomp;
if(/\s*\$[a-z]\w+\s*/i){
#if(/\b\$[a-z]\w+\b/i){
print Matched: $` -- $ -- $' :\n;
}
else{
print No match:$_\n;
}
}
##
##
I have another file that has this in it,
testfile.txt##
#
$he1lo is the name of a variable
what is not allowed $0 is this
but this is $a123wgfd343w cool
this is correct though $hello dont know why didnt work
before. I sure hope this doesnt pass alf$f12w32 cuz it
shouldnt $alfonso does match though $hello
##
The perl code is basically supposed to look for real possible scalar
variable names. So there should only be two failures,
the second and fifth line should not pass.
Now, if I run the program how it is, I get this,
answer1.txt###
##
Matched: -- $he1lo -- is the name of a variable :
No match:what is not allowed $0 is this
Matched: but this is -- $a123wgfd343w -- cool :
Matched: this is correct though -- $hello -- dont know why
didnt work
before. :
Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz
it shouldnt :
Matched: -- $alfonso -- does match though :
Matched: -- $hello -- :
No match:
##
if i run it with the commented section as the if statement,
and the current
if statement commented out, i get this,
###answer2.txt
##
No match:$he1lo is the name of a variable
No match:what is not allowed $0 is this
No match:but this is $a123wgfd343w cool
No match:this is correct though $hello dont know why didnt
work before.
Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz
it shouldnt : No match:$alfonso does match though No
match:$hello No match:
##
#
With regard to answer2.txt, why doesn't it match lines
1,3,4,5 and 6? In
the camel book(pg40), it says that the word bound \b matches
the beginning
of lines.
Any advice would be appreciated.
Thanks.
Beats me...I can't wait to find out the answer to this one. $Bill,
explain it to us idiots...
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Re: Another Regular expression problem
if(/\$\b[a-z]\w+\b/i)
should work. \b only includes those chars in \w which doesn't include
$. So $ is not include in the word, only hello is included.
ted zeng
Stovall, Adrian M. wrote:
-Original Message-
From: Electron One [mailto:[EMAIL PROTECTED]
Sent: Monday, March 10, 2003 12:12 PM
To: [EMAIL PROTECTED]
Cc: Electron One
Subject: Another Regular expression problem
Hello Everyone,
I have a perl file that has this,
PerlFile.pl###
#!/usr/bin/perl
while(){
chomp;
if(/\s*\$[a-z]\w+\s*/i){
#if(/\b\$[a-z]\w+\b/i){
print Matched: $` -- $ -- $' :\n;
}
else{
print No match:$_\n;
}
}
##
##
I have another file that has this in it,
testfile.txt##
#
$he1lo is the name of a variable
what is not allowed $0 is this
but this is $a123wgfd343w cool
this is correct though $hello dont know why didnt work
before. I sure hope this doesnt pass alf$f12w32 cuz it
shouldnt $alfonso does match though $hello
##
The perl code is basically supposed to look for real possible scalar
variable names. So there should only be two failures,
the second and fifth line should not pass.
Now, if I run the program how it is, I get this,
answer1.txt###
##
Matched: -- $he1lo -- is the name of a variable :
No match:what is not allowed $0 is this
Matched: but this is -- $a123wgfd343w -- cool :
Matched: this is correct though -- $hello -- dont know why
didnt work
before. :
Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz
it shouldnt :
Matched: -- $alfonso -- does match though :
Matched: -- $hello -- :
No match:
##
if i run it with the commented section as the if statement,
and the current
if statement commented out, i get this,
###answer2.txt
##
No match:$he1lo is the name of a variable
No match:what is not allowed $0 is this
No match:but this is $a123wgfd343w cool
No match:this is correct though $hello dont know why didnt
work before.
Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz
it shouldnt : No match:$alfonso does match though No
match:$hello No match:
##
#
With regard to answer2.txt, why doesn't it match lines
1,3,4,5 and 6? In
the camel book(pg40), it says that the word bound \b matches
the beginning
of lines.
Any advice would be appreciated.
Thanks.
Beats me...I can't wait to find out the answer to this one. $Bill,
explain it to us idiots...
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Re: Another Regular expression problem
At 12:06 PM 3/10/2003 -0800, Ted Zeng wrote:
if(/\$\b[a-z]\w+\b/i)
should work.
If i do this, it will still include this line,
I sure hope this doesnt pass alf$f12w32 cuz it shouldnt
And I want it to be a word that starts off with the $ symbol. :-(
\b only includes those chars in \w which doesn't include
$. So $ is not include in the word, only hello is included.
Can you explain this a little further, I dont understand you very well.
Thanks very much.
ted zeng
Stovall, Adrian M. wrote:
-Original Message-
From: Electron One [mailto:[EMAIL PROTECTED]
Sent: Monday, March 10, 2003 12:12 PM
To: [EMAIL PROTECTED]
Cc: Electron One
Subject: Another Regular expression problem
Hello Everyone,
I have a perl file that has this,
PerlFile.pl###
#!/usr/bin/perl
while(){
chomp;
if(/\s*\$[a-z]\w+\s*/i){
#if(/\b\$[a-z]\w+\b/i){
print Matched: $` -- $ -- $' :\n;
}
else{
print No match:$_\n;
}
}
##
##
I have another file that has this in it,
testfile.txt##
#
$he1lo is the name of a variable
what is not allowed $0 is this
but this is $a123wgfd343w cool
this is correct though $hello dont know why didnt work
before. I sure hope this doesnt pass alf$f12w32 cuz it
shouldnt $alfonso does match though $hello
##
The perl code is basically supposed to look for real possible scalar
variable names. So there should only be two failures,
the second and fifth line should not pass.
Now, if I run the program how it is, I get this,
answer1.txt###
##
Matched: -- $he1lo -- is the name of a variable :
No match:what is not allowed $0 is this
Matched: but this is -- $a123wgfd343w -- cool :
Matched: this is correct though -- $hello -- dont know why
didnt work
before. :
Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz
it shouldnt :
Matched: -- $alfonso -- does match though :
Matched: -- $hello -- :
No match:
##
if i run it with the commented section as the if statement,
and the current
if statement commented out, i get this,
###answer2.txt
##
No match:$he1lo is the name of a variable
No match:what is not allowed $0 is this
No match:but this is $a123wgfd343w cool
No match:this is correct though $hello dont know why didnt
work before.
Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz
it shouldnt : No match:$alfonso does match though No
match:$hello No match:
##
#
With regard to answer2.txt, why doesn't it match lines
1,3,4,5 and 6? In
the camel book(pg40), it says that the word bound \b matches
the beginning
of lines.
Any advice would be appreciated.
Thanks.
Beats me...I can't wait to find out the answer to this one. $Bill,
explain it to us idiots...
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RE: Another Regular expression problem
Electron One wrote, on Monday, March 10, 2003 13:12
: while(){
: chomp;
: if(/\s*\$[a-z]\w+\s*/i){
: #if(/\b\$[a-z]\w+\b/i){
: print Matched: $` -- $ -- $' :\n;
: }
: else{
:print No match:$_\n;
: }
: }
:
: testfile.txt##
: $he1lo is the name of a variable
: what is not allowed $0 is this
: but this is $a123wgfd343w cool
: this is correct though $hello dont know why didnt work before.
: I sure hope this doesnt pass alf$f12w32 cuz it shouldnt
: $alfonso does match though
: $hello
: ##
:
: Now, if I run the program how it is, I get this,
:
: Matched: -- $he1lo -- is the name of a variable :
: No match:what is not allowed $0 is this
: Matched: but this is -- $a123wgfd343w -- cool :
: Matched: this is correct though -- $hello -- dont know why didnt work
: before. :
: Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz it shouldnt :
: Matched: -- $alfonso -- does match though :
: Matched: -- $hello -- :
: No match:
: if i run it with the commented section as the if statement, and
: the current
: if statement commented out, i get this,
:
: No match:$he1lo is the name of a variable
: No match:what is not allowed $0 is this
: No match:but this is $a123wgfd343w cool
: No match:this is correct though $hello dont know why didnt work before.
: Matched: I sure hope this doesnt pass alf -- $f12w32 -- cuz it shouldnt :
: No match:$alfonso does match though
: No match:$hello
:
: Why doesn't the second match lines 1,3,4,5 and 6? In
: the camel book(pg40), it says that the word bound \b matches the
: beginning of lines.
The \b word boundary marks a boundary between a \w character and either
a \W character or one end of the line. I think what you really want is
this:
/(^|\W)\$[a-zA-Z]\w*\b/
Since $ is not a word character, you *do* get a word boundary when the $
appears in the middle of a word (hence your match on alf$f12w32 and nowhere
else). What you really want is to have the $ preceded by either the beginning
of the line or a non-word character. I also bet you *do* want to allow one-
character variables? If not, change the * back to a +.
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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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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Regular expression question
I want to translate some chaarctersin a string to its replacement string. I am doing this. #** $inchar = "\xa2\xa3";$outchar = "\[Cent]\[Pound]"; eval "\$buf =~ tr/$inchar/$outchar/"; #$buf =~ s/$inchar/$outchar/; print "$buf \n"; #** This is not working. Please Help. Thanks, -Mangesh
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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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
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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