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Re: How regex works in array mode?

by JavaFan (Canon)
on Apr 26, 2012 at 07:06 UTC ( #967253=note: print w/ replies, xml ) Need Help??


in reply to How regex works in array mode?

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.


Comment on Re: How regex works in array mode?
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...
      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})
      with
      [0-9]{1,3}
      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:
        ip:192.168.243.1

        but it doesn't print the "ip:192.168.243.2" Is it possible to print the next value in the array? Thanks

        or is this output possible?

        ip:192.168.243.1 ip:192.168.243.2

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