pattern matching

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks, I am familar with pattern matching but not sure how to tackle the following. I have a single line with numerous entries req=XX, where XX is a number.

req=44...something...req=56...req=24
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i need to extract each of numbers and print to a new file. I have tried the below put this only finds the first occurance then dies?

while (<>) {

if (m/req\=(\d+){1,400}/) {print OUT "$1\n"};
}
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Re: pattern matching by ikegami (Patriarch) on Jun 27, 2005 at 14:43 UTC
You need to use /g: `while (/req=(\d+)/g) { print OUT ("$1\n"); }` [download]	[reply] [d/l]
Re^2: pattern matching by Anonymous Monk on Jun 27, 2005 at 14:48 UTC
thanks for that, works fine cheers for the quick reponse	[reply]
Re: pattern matching by Ido (Hermit) on Jun 27, 2005 at 14:45 UTC
You should look at the /g modifier of the pattern matching operators: `while(m/.../g){ ...# Every match }` [download] BTW: What do you mean by (\d+){1,400}? It's something like "One or more digit, 1-400 times..", which makes the {1,400} somewhat redundant.	[reply] [d/l]
Re: pattern matching by JediWizard (Deacon) on Jun 27, 2005 at 14:48 UTC
To expand on other's answers with a link to the documentation: See perlop (look at the "Regexp Quote-Like Operators" section). They say that time changes things, but you actually have to change them yourself. —Andy Warhol	[reply] [d/l]
Re: pattern matching by GrandFather (Saint) on Jun 28, 2005 at 05:28 UTC
How about a "one liner": `use strict; print join ("\n", /req=(\d+)/g) . "\n" while <DATA>; __DATA__ req=44...something...req=56...req=24 req=23 diddle diddle diddle req=1 req=2` [download] Read more... Output (222 Bytes) Perl is Huffman encoded by design.	[reply] [d/l] [select]