Re^3: Question on Regex grouping

Numbering of captures counts per match/regex, of which you have two here - the one with the 8-digit capture being evaluated last.

If you wanted to extract the 5-digit number from the first regex, you could simply reverse the order of the &&-combined tests:

if (($arr =~ / (\w*)(def)(\d{8})/) && ($arr =~ /(\w*)(abc)(\d{5})/)) {
[download]

Now, $3 is the 5-digit number.

If you wanted to keep all captures, you could assign them to variables, e.g.:

if (( my ($c1, $c2, $c3) = $arr =~ /(\w*)(abc)(\d{5})/)
 && ( my ($c4, $c5, $c6) = $arr =~ / (\w*)(def)(\d{8})/)) {
[download]

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Re^4: Question on Regex grouping by ajguitarmaniac (Sexton) on Dec 21, 2010 at 07:05 UTC
Great! Just what I was looking for! Can I use the below mentioned approach to arrive at the same result? `if ($arr =~ /(\w)(abc)(\d{5})/){ $a = $3 if (($arr =~ / (\w)(def)(\d{8})/){ $b = $3 } } print $a; prnit $b;` [download]	[reply] [d/l]
Re^5: Question on Regex grouping by Anonyrnous Monk (Hermit) on Dec 21, 2010 at 07:12 UTC
Yes. But don't use `$a` and `$b` :) — they're special global variables (for sort).	[reply] [d/l] [select]
Re^6: Question on Regex grouping by ajguitarmaniac (Sexton) on Dec 21, 2010 at 07:40 UTC
Alright!! Thank you!! Let me see how sort works now :-)	[reply]
Re^5: Question on Regex grouping by AnomalousMonk (Archbishop) on Dec 22, 2010 at 05:38 UTC
Also remember that if either (or both) or the regexes fail to match, one or the other (or both) of `$a` and `$b` (or whatever you finally decide to call them) will be undefined.	[reply] [d/l] [select]