Re^3: How regex works in array mode?

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.

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Re^4: How regex works in array mode? by astronogun (Sexton) on Apr 26, 2012 at 09:12 UTC
I already replace it and the output is: `ip:192.168.243.1` [download] 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` [download]	[reply] [d/l] [select]
Re^5: How regex works in array mode? by JavaFan (Canon) on Apr 26, 2012 at 10:02 UTC
Perhaps you haven't replaced it correctly? `use 5.010; my @lines = ("ip:192.168.243.1", "ip:192.168.243.2" =~ /(ip:[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})/); say for @lines; __END__ ip:192.168.243.1 ip:192.168.243.2` [download] Do note that `ip:192.168.243.2` is in @lines only because the pattern matches the entire string. And do note that `"ip:192.168.243.1"` is not subject to any matching. In fact, the assignment to @lines is equivalent with: `my @lines; $lines[0] = "ip:192.168.243.1"; push @lines, $1 if "ip:192.168.243.2" =~ /(ip:[0-9]{1,3}\.[0-9]{1,3}\. +[0-9]{1,3}\.[0-9]{1,3})/;` [download]	[reply] [d/l] [select]