Re: how to next line after the pattern in perl

A simple state machine will work:

use strict;
use warnings;

my $flag = 0;
while (<DATA>) {
    if (/XXXX/) {
        $flag = 1;
    }
    elsif ($flag) {
        print;
        $flag = 0;
    }
}

__DATA__
ZZZZZZZZZZZZ
YYYYY
XXXX
XX
XXXX
3333
ii
XXXX
asdfddd
[download]

Comment on Re: how to next line after the pattern in perl Download Code

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Re^2: how to next line after the pattern in perl by kennethk (Abbot) on Jun 09, 2010 at 19:22 UTC
And a version that will catch consecutive lines and is shorter. Shorter is always better, right? `use strict; use warnings; my $flag = 0; while (<DATA>) { print if $flag; $flag = /XXXX/; } __DATA__ ZZZZZZZZZZZZ YYYYY XXXX XX XXXX 3333 ii XXXX asdfddd` [download] Correct, but fundamentally a repeat of Re: how to print next line after the pattern in perl.	[reply] [d/l]
Re^3: how to next line after the pattern in perl by ww (Archbishop) on Jun 09, 2010 at 19:41 UTC
"...catch consecutive lines...? and Correct but fundamentally.... Interesting proposition which points up a (possible) ambiguity or shortcoming in the original problem statement: What should be done when a second consecutive matching line exists? Print it and the following line? (which yours does) Print just the non-matching line? Something else? Update: Fixed punct at the end of the first li from "?" to ")" while LOL at kennethk's reply	[reply]
Re^4: how to next line after the pattern in perl by kennethk (Abbot) on Jun 09, 2010 at 20:03 UTC
Perhaps an appropriate solution? `use strict; use warnings; my $flag = 0; while (<DATA>) { print if $flag; if ($flag and /XXXX/) { eval('$demons->fly_out($nose)'); } $flag = /XXXX/; } __DATA__ ZZZZZZZZZZZZ YYYYY XXXX XX XXXX 3333 ii XXXX asdfddd` [download]	[reply] [d/l]