How regex works in array mode?

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

Hi, Once again I would like to seek for help..

It's simple I would like to know how regex works in perl in array mode.

I have this code wherein it will display the word ip: and the url x.x.x.x I have two inputed in my @lines and both are should be matched.. but the problem is I'm getting different result. Here's a simple code that I'm currently testing right now.

use strict;
use warnings;

my @lines = ("ip:192.168.243.1", "ip:192.168.243.2" =~ /(ip:([0-9]{3}|
+[0-9]{2}|[0-9]{1})\.([0-9]{3}|[0-9]{2}|[0-9]{1})\.([0-9]{3}|[0-9]{2}|
+[0-9]{1})\.([0-9]{3}|[0-9]{2}|[0-9]{1}))/);

print "@lines\n";
[download]

and the output is:

ip:192.168.243.1 ip:192.168.243.2 192 168 243 2
[download]

Suppose to be the "ip:192.168.243.1 ip:192.168.243.2" part is the one what I want it to be except for the next one "192 168 243 2"

I tried putting m and the regex code but it will just display "ip:192.168.243.1" even if the other value in the array is also matched (ip:192.168.243.2)

I tested the regex via http://regexpal.com/ and its working fine (I think hehe!) Hope you could help me learn how this regex works :) thank you so much

Comment on How regex works in array mode? Select or Download Code

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Re: How regex works in array mode? by JavaFan (Canon) on Apr 26, 2012 at 07:06 UTC
The first value is just the first value of @lines. The rest are return values of the match -- which returns each subpattern matched by a set of () in list context. If you want only the first one returned, get rid off all the capturing parens except the outer ones.	[reply]
Re^2: How regex works in array mode? by astronogun (Sexton) on Apr 26, 2012 at 08:32 UTC
Hmm.. How will I get rid all the capturing parens? Sorry for the noob question. still learning this whole regex thing...	[reply]
Re^3: How regex works in array mode? by JavaFan (Canon) on Apr 26, 2012 at 08:37 UTC
How will I get rid all the capturing parens? An easy way, one that should also speed up your pattern, make it more understandable, and saves typing, is to replace `([0-9]{3}\|[0-9]{2}\|[0-9]{1})` [download] with `[0-9]{1,3}` [download] Of course, you could also use the Regexp::Common module: but be aware, unlike your pattern, the one in Regexp::Common rejects invalid IP addresses.	[reply] [d/l] [select]
Re^4: How regex works in array mode? by astronogun (Sexton) on Apr 26, 2012 at 09:12 UTC
Re^5: How regex works in array mode? by JavaFan (Canon) on Apr 26, 2012 at 10:02 UTC
Re: How regex works in array mode? by jwkrahn (Abbot) on Apr 26, 2012 at 08:42 UTC
There is no such thing as "array mode" in Perl. I believe you are refering to "list context"? `([0-9]{3}\|[0-9]{2}\|[0-9]{1})` could be written more simply as `[0-9]{1,3}` and it will also avoid the unnecessary capturing parentheses. `use strict; use warnings; my @lines = ("ip:192.168.243.1", "ip:192.168.243.2" =~ /(ip:([0-9]{3}\| +[0-9]{2}\|[0-9]{1})\.([0-9]{3}\| +[0-9]{2}\|[0-9]{1})\.([0-9]{3}\|[0-9]{2}\|[0-9]{1})\.([0-9]{3}\|[0-9]{2}\| +[0-9]{1}))/); print "@lines\n";` [download] It looks like you want something like: `use strict; use warnings; my @lines = ( "ip:192.168.243.1", "ip:192.168.243.2" ); for ( @lines ) { print "$1\n" if /(ip:[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3 +})/; }` [download]	[reply] [d/l] [select]
Re^2: How regex works in array mode? by astronogun (Sexton) on Apr 26, 2012 at 10:14 UTC
Hi, Yes that's the output what i want it to be, you got me thanks :) My next question is.. is the 2 value on the array stored at $1? What if I want to use the value the "ip:192.168.243.2" how can i stored that at the other string for example $2 or something. Thanks Or something like if the items in the array are matched with the regex (like the code that you made. It will stored at $ipline1 = ip:192.168.243.1 and $ipline2 = ip:192.168.243.2 So that I can choose which one to output or print... Is it possible? Thanks	[reply]
Re: How regex works in array mode? by brx (Pilgrim) on Apr 26, 2012 at 09:09 UTC
`#!perl my $s = "how regex works in list context?"; my @capt= $s =~ /(((.)..) (.(.)...) ...(.)(.) .. (.(.))..( )....(.).(. +)(.))/; # 123 4 5 6 7 8 9 10 11 12 + 13 # h ow r e gex wor k s in l i st cont e x t + ? unshift @capt,'nothing'; print @capt[1,10,2,13,13,13,10,8,6,5,10,12,3,9,7]; print "\n"; #print "$_: $capt[$_]\n" for (0..13); __END__ how regex works in list context? how??? like this` [download] Read: perlre ("Capture groups"). Each opening parens creates a new capture. Uncomment last line if you want to see each group	[reply] [d/l]