Re: help in regular expression
ta == timothy adigun 2teezp...@gmail.com writes: ta You can use unpack: ta$val = 11.0.56.1; ta$new_val=unpack(x5 A2,$val); # skip forward 6, get 2 ta print $new_val # print 56; unpack would be a poor choice if the number of digits in a field changes. pack/unpack are meant for use in fixed field records. uri -- Uri Guttman -- uri AT perlhunter DOT com --- http://www.perlhunter.com -- Perl Developer Recruiting and Placement Services - - Perl Code Review, Architecture, Development, Training, Support --- -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
Re: help in regular expression
ug == Uri Guttman u...@stemsystems.com writes: ug unpack would be a poor choice if the number of digits in a field ug changes. pack/unpack are meant for use in fixed field records. That was a bad assumption on my side, I only considered the problem at hand, thinking unpack will be faster that substr. Thanks, timothy
Re: help in regular expression
Hi Irfan, On Sat, 10 Sep 2011 10:23:31 -0700 (PDT) Irfan Sayed irfan_sayed2...@yahoo.com wrote: hi, i have following string. $val = 11.0.56.1; i need to write regular expression which should match only 56 and print There are any number of ways to extract 56 using a regular expression from this string, and which would you would prefer depends on the general format of the strings like that that you expect. Can you describe how these strings look like? Please reply to the list. Regards, Shlomi Fish please suggest regards irfan -- - Shlomi Fish http://www.shlomifish.org/ Funny Anti-Terrorism Story - http://shlom.in/enemy Live as if you were to die tomorrow. Learn as if you were to live forever. — http://en.wikiquote.org/wiki/Mohandas_Gandhi (Disputed) Please reply to list if it's a mailing list post - http://shlom.in/reply . -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
Re: help in regular expression
On 11-09-10 01:23 PM, Irfan Sayed wrote: i have following string. $val = 11.0.56.1; i need to write regular expression which should match only 56 and print please suggest ( $val =~ /(56)/ ) print $1; -- Just my 0.0002 million dollars worth, Shawn Confusion is the first step of understanding. Programming is as much about organization and communication as it is about coding. The secret to great software: Fail early often. Eliminate software piracy: use only FLOSS. Make something worthwhile. -- Dear Hunter -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
Re: help in regular expression
On 10/09/2011 18:23, Irfan Sayed wrote: hi, i have following string. $val = 11.0.56.1; i need to write regular expression which should match only 56 and print please suggest I think you should forget about regular expressions and use split: my $sub = (split /\./, $val)[2]; HTH, Rob -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
Re: help in regular expression
Hi Irfan, You can use unpack: $val = 11.0.56.1; $new_val=unpack(x5 A2,$val); # skip forward 6, get 2 print $new_val # print 56;
Re: Help on regular expression !!
jet speed wrote:
Guys,
Hello,
I am new to perl, I am having trouble capturing the required output from
the command, with my limited knowlege i tried to put something togather. not
sure how to proceed beyond.
In a regular expression, when you want to capture part of a pattern you
have to enclose that part in parentheses.
What i am trying to achieve
for certain drives ex : B3494_901, B3494_102 from the outputlist is to find
the index number ex: 19 for drive B3494_102 as in the output below
then collect the different index number for these selected drives in a
variable $idx = 1:2:19:5
then i can use the $idx inside the command 'tpconfig -multiple_delete
-drive $idx '
Any help would be much appericiated.
Script
use strict;
use warnings;
my @tpd = `tpconfig -dl`;
my $idx;
my $drv;
foreach my $line (@tpd) {
chomp $line;
#$line =~ m/^Index\s+\d\d/ do {
$line =~ m/^Index\w+/ do {
It looks like the pattern you want is more like:
$line =~ /
^ # start of string
\s+ # followed by whitespace
Index # match literal string
\s+ # followed by whitespace
( \d+ ) # match and capture any numerical digits
/x
$idx = $1;
print $idx \n;
};
$line =~ /^Drive.*\s+\w\d+/ do {
$line =~ /
^ # start of string
\s+ # followed by whitespace
Drive\ Name # match literal string
\s+ # followed by whitespace
( \w+ ) # match and capture any word characters
/x
$drv =$1;
print $drv /n;
};
}
( tpconfig -dl )command output below
Drive Name B3494_102
Index 19
NonRewindDrivePath /dev/rmt/23cbn
Typehcart2
Status DOWN
SCSI Protection SR (Global)
Shared Access Yes
TLH(0) IBM Device Number=974680
Serial Number 07897468
Different Drives
Drive Name B3494_901
Drive Name B3494_100
Drive Name B3494_102
John
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Re: Help on regular expression !!
On Mon, Aug 3, 2009 at 4:00 PM, John W. Krahn jwkr...@shaw.ca wrote:
jet speed wrote:
Guys,
Hello,
I am new to perl, I am having trouble capturing the required output from
the command, with my limited knowlege i tried to put something togather.
not
sure how to proceed beyond.
In a regular expression, when you want to capture part of a pattern you
have to enclose that part in parentheses.
What i am trying to achieve
for certain drives ex : B3494_901, B3494_102 from the outputlist is to
find
the index number ex: 19 for drive B3494_102 as in the output below
then collect the different index number for these selected drives in a
variable $idx = 1:2:19:5
then i can use the $idx inside the command 'tpconfig -multiple_delete
-drive $idx '
Any help would be much appericiated.
Script
use strict;
use warnings;
my @tpd = `tpconfig -dl`;
my $idx;
my $drv;
foreach my $line (@tpd) {
chomp $line;
#$line =~ m/^Index\s+\d\d/ do {
$line =~ m/^Index\w+/ do {
It looks like the pattern you want is more like:
$line =~ /
^ # start of string
\s+ # followed by whitespace
Index # match literal string
\s+ # followed by whitespace
( \d+ ) # match and capture any numerical digits
/x
$idx = $1;
print $idx \n;
};
$line =~ /^Drive.*\s+\w\d+/ do {
$line =~ /
^ # start of string
\s+ # followed by whitespace
Drive\ Name # match literal string
\s+ # followed by whitespace
( \w+ ) # match and capture any word characters
/x
$drv =$1;
print $drv /n;
};
}
( tpconfig -dl )command output below
Drive Name B3494_102
Index 19
NonRewindDrivePath /dev/rmt/23cbn
Typehcart2
Status DOWN
SCSI Protection SR (Global)
Shared Access Yes
TLH(0) IBM Device Number=974680
Serial Number 07897468
Different Drives
Drive Name B3494_901
Drive Name B3494_100
Drive Name B3494_102
John
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Hi John, Thanks for your help, Much appreciated. Please could you also
refer any good reference for Regular expression for a beginer like me, would
be great.
Cheers, Sj
Re: Help on regular expression !!
jet speed wrote: Hi John, Thanks for your help, Much appreciated. Please could you also refer any good reference for Regular expression for a beginer like me, would be great. Have you read the documentation that comes with Perl? perldoc perlrequick perldoc perlretut perldoc perlre John -- Those people who think they know everything are a great annoyance to those of us who do.-- Isaac Asimov -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
Re: Help on regular expression !!
On Mon, Aug 3, 2009 at 5:45 PM, John W. Krahn jwkr...@shaw.ca wrote: jet speed wrote: Hi John, Thanks for your help, Much appreciated. Please could you also refer any good reference for Regular expression for a beginer like me, would be great. Have you read the documentation that comes with Perl? perldoc perlrequick perldoc perlretut perldoc perlre John -- Those people who think they know everything are a great annoyance to those of us who do.-- Isaac Asimov -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/ Hi John, Thanks again, I have been jumping between manuals orelly books mainly Programming 3rd edition Perl, cookbook etc.. I do find the orelly books hard to follow. I will give it a shot the documentation you have referred. Cheers, Sj
Re: Help on regular expression !!
jet speed wrote: On Mon, Aug 3, 2009 at 5:45 PM, John W. Krahn jwkr...@shaw.ca wrote: jet speed wrote: Hi John, Thanks for your help, Much appreciated. Please could you also refer any good reference for Regular expression for a beginer like me, would be great. Have you read the documentation that comes with Perl? perldoc perlrequick perldoc perlretut perldoc perlre Hi John, Thanks again, I have been jumping between manuals orelly books mainly Programming 3rd edition Perl, cookbook etc.. I do find the orelly books hard to follow. I will give it a shot the documentation you have referred. Cheers, Sj The best reference (IMHO) is the book from O'Reilly: http://oreilly.com/catalog/9780596528126/ John -- Those people who think they know everything are a great annoyance to those of us who do.-- Isaac Asimov -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
Re: Help on regular expression
--- Sayed, Irfan (Irfan) [EMAIL PROTECTED] wrote: Hi All, I have a string like this CLEARCASE_CMDLINE = (mkact -nc notme sprint) Now with the regular expression what I want is only those characters before closing braces excluding white space character. I mean to say that if regular expression encounter the white space character then it should stop. So my output should come as sprint. What result string do you want? for example? Best Regards, Jeff (joy) Peng Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Help on regular expression
On Dec 13, 2007 6:12 AM, Sayed, Irfan (Irfan) [EMAIL PROTECTED] wrote: Hi All, I have a string like this CLEARCASE_CMDLINE = (mkact -nc notme sprint) Now with the regular expression what I want is only those characters before closing braces excluding white space character. I mean to say that if regular expression encounter the white space character then it should stop. So my output should come as sprint. Regards Irfan. What have you tried? What about it is confusing you? -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Help on regular expression
Sayed, Irfan (Irfan) wrote: Hi All, I have a string like this CLEARCASE_CMDLINE = (mkact -nc notme sprint) Now with the regular expression what I want is only those characters before closing braces excluding white space character. I mean to say that if regular expression encounter the white space character then it should stop. So my output should come as sprint. Does the program below do what you need? HTH, Rob use strict; use warnings; my $str = 'CLEARCASE_CMDLINE = (mkact -nc notme sprint)'; $str =~ /(\S+)\s*\)/ or die String didn't match expected pattern; my $word = $1; print Word found is '$word'\n; -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
RE: Help on regular expression
Thanks Rob. It really helped. I got what I want. Request you to please explain me in detail how you did this. I did not understand much. Please guide. Regards Irfan. -Original Message- From: Rob Dixon [mailto:[EMAIL PROTECTED] Sent: Thursday, December 13, 2007 6:21 PM To: beginners@perl.org Perl Beginners Cc: Sayed, Irfan (Irfan) Subject: Re: Help on regular expression Sayed, Irfan (Irfan) wrote: Hi All, I have a string like this CLEARCASE_CMDLINE = (mkact -nc notme sprint) Now with the regular expression what I want is only those characters before closing braces excluding white space character. I mean to say that if regular expression encounter the white space character then it should stop. So my output should come as sprint. Does the program below do what you need? HTH, Rob use strict; use warnings; my $str = 'CLEARCASE_CMDLINE = (mkact -nc notme sprint)'; $str =~ /(\S+)\s*\)/ or die String didn't match expected pattern; my $word = $1; print Word found is '$word'\n; -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Help on regular expression
Sayed, Irfan (Irfan) wrote: From: Rob Dixon [mailto:[EMAIL PROTECTED] Sayed, Irfan (Irfan) wrote: Hi All, I have a string like this CLEARCASE_CMDLINE = (mkact -nc notme sprint) Now with the regular expression what I want is only those characters before closing braces excluding white space character. I mean to say that if regular expression encounter the white space character then it should stop. So my output should come as sprint. Does the program below do what you need? use strict; use warnings; my $str = 'CLEARCASE_CMDLINE = (mkact -nc notme sprint)'; $str =~ /(\S+)\s*\)/ or die String didn't match expected pattern; my $word = $1; print Word found is '$word'\n; Thanks Rob. It really helped. I got what I want. Request you to please explain me in detail how you did this. I did not understand much. Please guide. OK - \S matches any non-space character, so \S+ matches a sequence of one or more of them - \s matches any space character, so \s* matches an optional sequence of them - \) matches a closing parenthesis - so the regular expression /\S+\s*\)/ matches a sequence of non-space characters, followed optionally by a some whitespace, followed by a closing parenthesis, which is what the relevant part of your string looks like - finally, putting parentheses around \S causes the string of non-space characters to be stored in $1 I hope that makes it clearer. Rob -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: help in regular expression
Irfan J Sayed wrote: Hi , Hello, I am using following code #!/usr/local/bin/perl # Main program use warnings; use strict; my $file = c:\backup.pl; The escape sequence \b represents the backspace character. You probably want: my $file = 'c:/backup.pl'; open(FH,$file) || die can't open a file; You should include the $file variable and the $! variable in the error message: open(FH,$file) || die can't open '$file' $!; my $pattern = '\w\s\w'; my $input = ; You probably want to get your input from the FH filehandle: my $input = FH; print yes got the match if $input =~ /$pattern/; but i am getting following error Name main::FH used only once: possible typo at C:\irfan\search.pl line 9. You are only using the FH filehandle once, in the open function. can't open a file at C:\irfan\search.pl line 9. You probably don't have a file name with a backspace character in it. John -- use Perl; program fulfillment -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: help in regular expression
Hi Irfan, On 6/1/06, Irfan J Sayed [EMAIL PROTECTED] wrote: Hi , I am using following code #!/usr/local/bin/perl # Main program use warnings; use strict; my $file = c:\backup.pl; open(FH,$file) || die can't open a file; my $pattern = '\w\s\w'; my $input = ; print yes got the match if $input =~ /$pattern/; but i am getting following error Name main::FH used only once: possible typo at C:\irfan\search.pl line 9. can't open a file at C:\irfan\search.pl line 9. can anybody plz help me out As far as I can tell, you need to escape the '\' in the string you're assigning to $file: my $file = c:\\backup.pl; You also need to either escape the slashes for the pattern you're using, or use the qr// form: my $pattern = qr/\w\s\w/; # or my $pattern = '\\w\\s\\w'; HTH, David -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
interpolation problems (was: Re: help in regular expression)
Irfan J Sayed schreef:
#!/usr/local/bin/perl
use warnings;
use strict;
Good!
my $file = c:\backup.pl;
Use
my $file = 'c:\backup.pl';
or rather
my $file = 'c:/backup.pl';
open(FH,$file) || die can't open a file;
Make that:
open my $fh, '', $file or die Error opening '$file': $! ;
my $pattern = '\w\s\w';
my $pattern = qr/\w\s\w/ ;
my $input = ;
print yes got the match if $input =~ /$pattern/;
Try this:
while ( $fh ) {
/$pattern/ and printf match in line %s\n, $. ;
}
but i am getting following error
Name main::FH used only once: possible typo at C:\irfan\search.pl
close $fh or die Error closing '$file': $! ;
--
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Re: help in regular expression
Irfan J Sayed wrote: Hi , I am using following code #!/usr/local/bin/perl # Main program use warnings; use strict; my $file = c:\backup.pl; open(FH,$file) || die can't open a file; [...] For the die statement, use this instead: die can't open this file: $file reason: $!; Your problem is that \b has a special meaning in a double-quoted string. It's a backspace character. Escape the backslash by putting another backslash before it. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: help in regular expression
David Romano schreef:
[ $pattern = '\w\s\w' ]
You also need to [...] escape the slashes for the pattern you're
using
I don't think that is needed:
(1) perl -le '$re = q{\w\s\w} ; print qr/$re/'
(2) perl -le '$re = q{\\w\\s\\w} ; print qr/$re/'
(on Windows you'll need to change the outer quotes to dquotes)
Both will print: (?-xism:\w\s\w).
Try also:
perl -le 'print q{\\w\\s\\w}'
perl -le 'print q{\w\s\w}'
that both print: \w\s\w.
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Re: help in regular expression
Hi Ruud,
On 6/1/06, Dr.Ruud [EMAIL PROTECTED] wrote:
David Romano schreef:
[ $pattern = '\w\s\w' ]
You also need to [...] escape the slashes for the pattern you're
using
I don't think that is needed:
(1) perl -le '$re = q{\w\s\w} ; print qr/$re/'
(2) perl -le '$re = q{\\w\\s\\w} ; print qr/$re/'
(on Windows you'll need to change the outer quotes to dquotes)
Both will print: (?-xism:\w\s\w).
Try also:
perl -le 'print q{\\w\\s\\w}'
perl -le 'print q{\w\s\w}'
that both print: \w\s\w.
Yep, you're right. Looking at it again, my only excuse is that it was
4 in the morning. When I was writing that, I forgot what I learned
from the Llama book that only \' and \\ work between single-quotes
(thus the suggestion to change c:\backup.pl to c:\\backup.pl
instead of 'c:\backup.pl'). Thanks again for pointing out where I
erred :-)
David
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Re: help on regular expression
Madhu Reddy wrote:
Hi,
I need some help on regular expression...
i have following in variable $total_count
$total_count = ##I USBP 01 10:38:09(000)
xyz_abc_etrack_validation,6 ETRACK_TOTAL_RECS : 100
Here in this ETRACK_TOTAL_RECS is fixed and common for
all and rest is changing...
like following
$total_count = ##I USBP 02 12:38:09(000)
abc_gkkh_uiu,8 ETRACK_TOTAL_RECS : 500
here i need some regular expression to get, 100 like
following
$total_count = 100
I appreciate u r help..
Hi Madhu.
There are two separate things that regexes are good for:
verifying that a string matches a given pattern, and
extracting pieces of data from the string. Usually you want
a mixture of both at the same time. Code looking something
like this should do what you want:
if ($total_count =~ m/\sETRACK_TOTAL_RECS\s*:\s*(\d+)/) {
my $total_recs = $1;
print ETRACK_TOTAL_RECS = $total_recs\n;
}
else {
die Unexpected record format: $total_count;
}
But there are endless variations depending on how much checking
you want to build in.
HTH,
Rob
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RE: help on regular expression
Hi, I need some help on regular expression... i have following in variable $total_count $total_count = ##I USBP 01 10:38:09(000) xyz_abc_etrack_validation,6 ETRACK_TOTAL_RECS : 100 Here in this ETRACK_TOTAL_RECS is fixed and common for all and rest is changing... like following $total_count = ##I USBP 02 12:38:09(000) abc_gkkh_uiu,8 ETRACK_TOTAL_RECS : 500 here i need some regular expression to get, 100 like following $total_count = 100 tmtowtdi, but if I knew that I was going to have a number with no punctuation after the text ETRACK_TOTAL_RECS : that I wanted to capture, I'd do: $total_count =~ /ETRACK_TOTAL_RECS\s*:\s*(\d+)/; Then the number would be in $1. If I wasn't sure of the format (i.e., it could have a decimal, commas, etc), but knew that I wanted to capture everything from the : to the end of the string, I'd do: $total_count =~ /ETRACK_TOTAL_RECS\s*:\s*(.+)/; And the result would be in $1. Hope this helps, /\/\ark -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: help on regular expression
On Tue, Jan 27, 2004 at 07:49:02AM -0800, Madhu Reddy ([EMAIL PROTECTED]) wrote:
Hi,
I need some help on regular expression...
i have following in variable $total_count
$total_count = ##I USBP 01 10:38:09(000)
xyz_abc_etrack_validation,6 ETRACK_TOTAL_RECS : 100
Here in this ETRACK_TOTAL_RECS is fixed and common for
all and rest is changing...
like following
$total_count = ##I USBP 02 12:38:09(000)
abc_gkkh_uiu,8 ETRACK_TOTAL_RECS : 500
here i need some regular expression to get, 100 like
following
$total_count = 100
I appreciate u r help..
I'm not entirely sure what you are trying to do here but I'll take a
stab at it. I'm assuming you have lines in a file or database like -
##I USBP 01 10:38:09(000) xyz_abc_etrack_validation,6 ETRACK_TOTAL_RECS : 100
##I USBP 02 12:38:09(000) abc_gkkh_uiu,8 ETRACK_TOTAL_RECS : 500
You want to pull the last number from each of those lines and total them
up. If that is the case then the following example may help you. If my example
is what you're looking for and you need more explenation let me know. If it
isn't what you are looking for, explain a little more about what you are
trying to do and show us the code you have written.
#!/usr/bin/perl
use warnings;
use strict;
my $total_count;
while(DATA) {
$_ =~ /:\s+(\d+)/;
$total_count += $1;
}
print $total_count\n;
__DATA__
##I USBP 01 10:38:09(000) xyz_abc_etrack_validation,6 ETRACK_TOTAL_RECS : 100
##I USBP 02 12:38:09(000) abc_gkkh_uiu,8 ETRACK_TOTAL_RECS : 500
Kent
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Re: help on regular expression
Hi, $total_count =~ s/.+ : (\d+)/$1/; HTH, Jan Madhu Reddy wrote: Hi, I need some help on regular expression... i have following in variable $total_count $total_count = ##I USBP 01 10:38:09(000) xyz_abc_etrack_validation,6 ETRACK_TOTAL_RECS : 100 Here in this ETRACK_TOTAL_RECS is fixed and common for all and rest is changing... like following $total_count = ##I USBP 02 12:38:09(000) abc_gkkh_uiu,8 ETRACK_TOTAL_RECS : 500 here i need some regular expression to get, 100 like following $total_count = 100 I appreciate u r help.. Thanks -Madhu __ Do you Yahoo!? Yahoo! SiteBuilder - Free web site building tool. Try it! http://webhosting.yahoo.com/ps/sb/ -- A common mistake that people make when trying to design something completely foolproof is to underestimate the ingenuity of complete fools. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: help on regular expression
On Tue, 27 Jan 2004, Madhu Reddy wrote: Hi, I need some help on regular expression... i have following in variable $total_count $total_count = ##I USBP 01 10:38:09(000) xyz_abc_etrack_validation,6 ETRACK_TOTAL_RECS : 100 Here in this ETRACK_TOTAL_RECS is fixed and common for all and rest is changing... like following $total_count = ##I USBP 02 12:38:09(000) abc_gkkh_uiu,8 ETRACK_TOTAL_RECS : 500 here i need some regular expression to get, 100 like following $total_count = 100 I appreciate u r help.. Thanks -Madhu From your examples, it appears that the number you want is always at the end of the line. The simpliest, but not necessarily the best way to do this might be: ($total_count) = ($total_count =~ /(\d+)$/); Basically, this says: Match all the digits before the end of the line. -- Maranatha! John McKown -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: Help with regular expression
Try this if you want regex.
This may not be the appropriate but works.
@list = ('1992','1993 (summer)','1995 fall');
foreach (@list) {
push @array, $1 if $_ =~ /(\d+)\s*.*$/;
}
print map {$_,\n} @array;
regards
Rajendran
Burlingame,CA
- Original Message -
From: Shaun Bramley [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, September 25, 2002 1:24 PM
Subject: Help with regular expression
Hi all,
I'm just looking for some confirmation on my regx.
I have a list ('1992', '1993 (summer)', '1995 fall') and what I want to do
is keep only the first four characters. Will the regx work for me?
@list =~ s/\s*//;
Again will that turn the list into (1992, 1993, 1995)?
as always thank you for your time and effort in helping me
Shaun Bramley
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RE: Help with regular expression
Nope..this won't work. Why don't you loop over the list and do a substring or pack as
you know that you need to keep only the first 4 characters of each element?
-Original Message-
From: Shaun Bramley [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, September 25, 2002 4:24 PM
To: [EMAIL PROTECTED]
Subject: Help with regular expression
Hi all,
I'm just looking for some confirmation on my regx.
I have a list ('1992', '1993 (summer)', '1995 fall') and what I want to do
is keep only the first four characters. Will the regx work for me?
@list =~ s/\s*//;
Again will that turn the list into (1992, 1993, 1995)?
as always thank you for your time and effort in helping me
Shaun Bramley
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RE: Help with regular expression
-Original Message-
From: Shaun Bramley [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, September 25, 2002 4:24 PM
To: [EMAIL PROTECTED]
Subject: Help with regular expression
Hi all,
I'm just looking for some confirmation on my regx.
I have a list ('1992', '1993 (summer)', '1995 fall') and what
I want to do
is keep only the first four characters. Will the regx work for me?
@list =~ s/\s*//;
Again will that turn the list into (1992, 1993, 1995)?
No.
1. You can't have a list on the left side of =~. To apply the substitution
over the list, use:
s/\s*// for @list;
2. That regex will strip leading whitespace from the string, if any.
To grab just the first 4 chars, you don't even need a regex:
$_ = substr($_, 0, 4) for $list;
HTH
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Re: Help with regular expression
Shaun Bramley wrote:
Hi all,
I'm just looking for some confirmation on my regx.
I have a list ('1992', '1993 (summer)', '1995 fall') and what I want to do
is keep only the first four characters. Will the regx work for me?
@list =~ s/\s*//;
Again will that turn the list into (1992, 1993, 1995)?
you can use substr() to null out from the 5th byte:
substr($_,4) = '' for(@list);
or you can use the unpack() function which could be a bit faster:
print unpack('A4',$_),\n for(@list);
david
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Re: Help with regular expression
Shaun Bramley wrote at Wed, 25 Sep 2002 22:24:00 +0200:
I'm just looking for some confirmation on my regx.
I have a list ('1992', '1993 (summer)', '1995 fall') and what I want to do
is keep only the first four characters. Will the regx work for me?
@list =~ s/\s*//;
Again will that turn the list into (1992, 1993, 1995)?
Another than using substr is really to use a regexp:
my @year = map /(\d\d\d\d)/, @list;
I would prefer it against the substr solution,
as it better express what you want to get out of the string.
(4 digits representing a year).
It has the disadvantage that it is slower than the substr solution.
Greetings,
Janek
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Re: Help in Regular expression with array
At 15:49 2002.06.05, Ankit Gupta wrote:
Hello,
I am facing a problem in using regular expression on array. My code is
written below:
open(FILE, $dirvalue) ;
my @lines = FILE;
print @lines; # prints the file contents
if( @lines =~ m/Date:/) { print ok;}
close(FILE);
here I can print @lines which gives me:
the =~ operator works on a scalar, not an array. You'll have to wait for Perl 6 to do
that :-).
Received: from pc_jrs.spacebel.be (pc_jr) by earth.spacebel (5.x/SMI-SVR4)
id AA26092; Sun, 27 Oct 1996 16:44:52 +0100
Date: Sun, 27 Oct 96 17:38:51 PST
From: John Reynolds [EMAIL PROTECTED]
Subject: Ebnf2ps
Now @lines does contain Date: characters but it does not return true
anwswer. Could someone please help me as how I can achieve this.
Here's another way to do the same (warning, I didn't syntax check the code).
open FILE, $dirvalue or die Can't open $dirvalue: $!;
my $found_date = 0;
while(FILE) {
print; # This print $_ which is the current line
$found_date = 1 if /Date:/; # the // is applied to $_ by default
}
close FILE or die Can't close $dirvalue: $!;
print OK\n if $found_date;
Hope this helps
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RE: Help in Regular expression with array
-Original Message-
From: Ankit Gupta [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 05, 2002 3:49 PM
To: [EMAIL PROTECTED]
Subject: Help in Regular expression with array
Hello,
I am facing a problem in using regular expression on array. My code is
written below:
open(FILE, $dirvalue) ;
my @lines = FILE;
print @lines; # prints the file contents
Here @lines is used in list context, so you get the contents of
the array.
if( @lines =~ m/Date:/) { print ok;}
Here @lines is used in scalar context, so you get a number representing
the number of elements in the array.
close(FILE);
here I can print @lines which gives me:
Received: from pc_jrs.spacebel.be (pc_jr) by earth.spacebel
(5.x/SMI-SVR4)
id AA26092; Sun, 27 Oct 1996 16:44:52 +0100
Date: Sun, 27 Oct 96 17:38:51 PST
From: John Reynolds [EMAIL PROTECTED]
Subject: Ebnf2ps
Now @lines does contain Date: characters but it does not return true
anwswer. Could someone please help me as how I can achieve this.
You need a loop:
/Date:/ print(ok\n), last for @lines;
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RE: Help in Regular expression with array
Hi -
I don't think you can regex on a whole array; try this
after you have loaded the array:
for (@lines) {print ok if /Date:/; }
This iterates the array lines presenting $_ for each
iteration. The regex /Date:/ operates on $_.
Aloha = Beau.
-Original Message-
From: Ankit Gupta [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 05, 2002 9:49 AM
To: [EMAIL PROTECTED]
Subject: Help in Regular expression with array
Hello,
I am facing a problem in using regular expression on array. My code is
written below:
open(FILE, $dirvalue) ;
my @lines = FILE;
print @lines; # prints the file contents
if( @lines =~ m/Date:/) { print ok;}
close(FILE);
here I can print @lines which gives me:
Received: from pc_jrs.spacebel.be (pc_jr) by earth.spacebel (5.x/SMI-SVR4)
id AA26092; Sun, 27 Oct 1996 16:44:52 +0100
Date: Sun, 27 Oct 96 17:38:51 PST
From: John Reynolds [EMAIL PROTECTED]
Subject: Ebnf2ps
Now @lines does contain Date: characters but it does not return true
anwswer. Could someone please help me as how I can achieve this.
Thanx
Ankit
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RE: Help in Regular expression with array
Don't know if you can do a search on an array (until and unless you want to evaluate
each element)
In case you are trying to achieve the multiple lines search, maybe this or the 2nd
example can help :
open (FILE , $ARGV[0]);
my @lines = FILE;
close(FILE);
$line = join( , @lines);
print $line;
if( $line =~ /Date:/m) { print ok;}
---
else you may try this way in case you have only one file (note)
---
while () {
print ok if (/Date:/);
}
else the old way or looping over each element
open (FILE , $ARGV[0]);
print ok if ( map { /Date:/ } (FILE) );
close FILE;
-Original Message-
From: Ankit Gupta [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 05, 2002 3:49 PM
To: [EMAIL PROTECTED]
Subject: Help in Regular expression with array
Hello,
I am facing a problem in using regular expression on array. My code is
written below:
open(FILE, $dirvalue) ;
my @lines = FILE;
print @lines; # prints the file contents
if( @lines =~ m/Date:/) { print ok;}
close(FILE);
here I can print @lines which gives me:
Received: from pc_jrs.spacebel.be (pc_jr) by earth.spacebel (5.x/SMI-SVR4)
id AA26092; Sun, 27 Oct 1996 16:44:52 +0100
Date: Sun, 27 Oct 96 17:38:51 PST
From: John Reynolds [EMAIL PROTECTED]
Subject: Ebnf2ps
Now @lines does contain Date: characters but it does not return true
anwswer. Could someone please help me as how I can achieve this.
Thanx
Ankit
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RE: Help in Regular expression with array
At 16:12 2002.06.05, Shishir K. Singh wrote:
open (FILE , $ARGV[0]);
print ok if ( map { /Date:/ } (FILE) );
close FILE;
map return an array with the result of the express apply to each line. Even if none of
the lines in FILE contain Date:, you will have an array with one value for each
line. This is a non empty array i.e. it has multiple lines and when evaluate in a
scalar context (with the if() ), it would always be true unless there are no lines in
FILE.
I think you want grep there as in
print ok if (grep /Date:/ ,(FILE));
or
print ok if (map { /Date:/ ? $_ : () } , (FILE));
I may be wrong though.
Best
--
Éric Beaudoin mailto:[EMAIL PROTECTED]
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RE: Help in Regular expression with array
Beau..I guess , the evaluation of the expression is not going to be true if no
Date: is found. Since map returns a list consisting of the results of each
successive evaluation of the expression..., the map will return undef.
Although, here..I think using grep would be a better idea!!
-Original Message-
From: Eric Beaudoin [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 05, 2002 4:29 PM
To: Shishir K. Singh
Cc: Ankit Gupta; [EMAIL PROTECTED]
Subject: RE: Help in Regular expression with array
At 16:12 2002.06.05, Shishir K. Singh wrote:
open (FILE , $ARGV[0]);
print ok if ( map { /Date:/ } (FILE) );
close FILE;
map return an array with the result of the express apply to each line. Even if none of
the lines in FILE contain Date:, you will have an array with one value for each
line. This is a non empty array i.e. it has multiple lines and when evaluate in a
scalar context (with the if() ), it would always be true unless there are no lines in
FILE.
I think you want grep there as in
print ok if (grep /Date:/ ,(FILE));
or
print ok if (map { /Date:/ ? $_ : () } , (FILE));
I may be wrong though.
Best
--
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RE: Help in Regular expression with array
At this point I feel compelled to mention that if the file you will be
working on is more than a few K or so, then it might be a good idea to
rewrite your loop so that you don't dump the entire file to memory.
Something like:
open(FILE,myfile.txt);
while(FILE){
if($_ =~ /Date:/){
print Ok!\n;
DoSomeOtherStuff();
}
}
Which you'll notice is already being done in some form in a few of the
responses.
-Original Message-
From: Shishir K. Singh [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 05, 2002 1:48 PM
To: Eric Beaudoin
Cc: Ankit Gupta; [EMAIL PROTECTED]
Subject: RE: Help in Regular expression with array
Beau..I guess , the evaluation of the expression is not going to be true if
no Date: is found. Since map returns a list consisting of the results of
each successive evaluation of the expression..., the map will return undef.
Although, here..I think using grep would be a better idea!!
-Original Message-
From: Eric Beaudoin [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 05, 2002 4:29 PM
To: Shishir K. Singh
Cc: Ankit Gupta; [EMAIL PROTECTED]
Subject: RE: Help in Regular expression with array
At 16:12 2002.06.05, Shishir K. Singh wrote:
open (FILE , $ARGV[0]);
print ok if ( map { /Date:/ } (FILE) );
close FILE;
map return an array with the result of the express apply to each line. Even
if none of the lines in FILE contain Date:, you will have an array with
one value for each line. This is a non empty array i.e. it has multiple
lines and when evaluate in a scalar context (with the if() ), it would
always be true unless there are no lines in FILE.
I think you want grep there as in
print ok if (grep /Date:/ ,(FILE));
or
print ok if (map { /Date:/ ? $_ : () } , (FILE));
I may be wrong though.
Best
--
Éric Beaudoin mailto:[EMAIL PROTECTED]
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RE: help in regular expression
You forgot to add g (global)in the end... $dirstruct =~ s/([\W])/-/g; Cheers Shishir -Original Message- From: Ankit Gupta [mailto:[EMAIL PROTECTED]] Sent: Friday, May 31, 2002 10:53 AM To: [EMAIL PROTECTED] Subject: help in regular expression Hello Friends, I need help in the below written script. $dirstruct =~ s/([\W])/-/; print $dirstruct; here $dirstruct is c:\ankit\test what I need as output is c--ankit-test but the output given by my script is c-\ankit\test Thanx Ankit -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
Re: help in regular expression
On May 31, Ankit Gupta said: $dirstruct =~ s/([\W])/-/; You're missing the /g modifier. And s/([\W])/-/g could be written as s/\W/-/g and would be a bit more efficient. -- Jeff japhy Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/ RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/ ** Look for Regular Expressions in Perl published by Manning, in 2002 ** stu what does y/// stand for? tenderpuss why, yansliterate of course. [ I'm looking for programming work. If you like my work, let me know. ] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
RE: Help with regular expression.
You are looking for the word addr: followed by four groups of digits, three of which are separated by a dot. Therefore you want $string =~ /addr:(\d+\.\d+.\d+\.\d+)/; Now, $1 has the correct ip address. -Original Message- From: Daniel Falkenberg To: [EMAIL PROTECTED] Sent: 10/4/2001 6:43 PM Subject: Help with regular expression. List, I have an IP address within this regular expression that I need extracting and stored in a variable. Could some one offer some help on this? The line is as follows... inet addr:144.137.215.25 P-t-P:172.31.28.24 Mask:255.255.255.255 I need to extract the inet addr: i.e 144.137.215.25 and store it in a variable. I am very new to regexes. So any help would be greatly appriciated. Daniel Falkenberg. == VINTEK CONSULTING PTY LTD (ACN 088 825 209) Email: [EMAIL PROTECTED] WWW:http://www.vintek.net Tel:(08) 8523 5035 Fax:(08) 8523 2104 Snail: P.O. Box 312 Gawler SA 5118 == -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
