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Re^2: How regex works in array mode?

by astronogun (Sexton)
on Apr 26, 2012 at 08:32 UTC ( #967266=note: print w/replies, xml ) Need Help??

in reply to Re: How regex works in array mode?
in thread How regex works in array mode?

Hmm.. How will I get rid all the capturing parens? Sorry for the noob question. still learning this whole regex thing...

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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
    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.
      I already replace it and the output is:

      but it doesn't print the "ip:" Is it possible to print the next value in the array? Thanks

      or is this output possible?

      ip: ip:
        Perhaps you haven't replaced it correctly?
        use 5.010; my @lines = ("ip:", "ip:" =~ /(ip:[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})/); say for @lines; __END__ ip: ip:
        Do note that ip: is in @lines only because the pattern matches the entire string. And do note that "ip:" is not subject to any matching. In fact, the assignment to @lines is equivalent with:
        my @lines; $lines[0] = "ip:"; push @lines, $1 if "ip:" =~ /(ip:[0-9]{1,3}\.[0-9]{1,3}\. +[0-9]{1,3}\.[0-9]{1,3})/;

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