http://www.perlmonks.org?node_id=557535

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

This node falls below the community's threshold of quality. You may see it by logging in.

Replies are listed 'Best First'.
Re: Counting the number of replaces
by McDarren (Abbot) on Jun 26, 2006 at 09:42 UTC
    Yes, just evaluate your regexp in scalar context, eg: 
    $num_replacements = $_ =~ s/<I>/<italic>/g;

    [download]

    Cheers,
    Darren :)

[reply]
[d/l]
Re: Counting the number of replaces
by GrandFather (Saint) on Jun 26, 2006 at 09:46 UTC

    Note too that the $_ =~ is redundant. The regex on its own is quite sufficient: my $count = s/<I>/<italic>/g;.

    DWIM is Perl's answer to Gödel
[reply]
[d/l]
[select]
Re: Counting the number of replaces
by pingo (Hermit) on Jun 26, 2006 at 10:11 UTC
    In the spirit of TIMTOWTDI: 
    s/<I>/$count++;"<italic>"/eg;
print $count || 0;

    [download]
    No, I didn't say it was pretty. ;-)
[reply]
[d/l]
Re: Counting the number of replaces
by cognizant (Pilgrim) on Jun 26, 2006 at 09:38 UTC
    Try this. $a = $b=~s/<i>/<italic>/g; print "$a"; Gives the output you need.
[reply]
[d/l]
[select]
      I know this is only a quick example, but in general i would avoid using the special variables $a and $b


      This is not a Signature...
[reply]
Re: Counting the number of replaces
by Moron (Curate) on Jun 26, 2006 at 13:49 UTC
    If you want to get the number of replacements made in the whole file, not just for a particular line, then these should accumulate e.g. using += : 
    while( <$inputfh> ) {
    $replcount += s/replacethis/withthis/g;
    print $outputfh $_;
}

    [download]

    -M

    Free your mind

[reply]
[d/l]
Re: Counting the number of replaces
by shmem (Chancellor) on Jun 26, 2006 at 09:54 UTC
    rsriram, please read How do I post a question effectively?. Your could have found the answer for your last questions by yourself, skimming through perfunc (perl function reference) and perlop (perl operator reference).

    cheers,
    --shmem 

    _($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}
[reply]
Re: Counting the number of replaces
by ForgotPasswordAgain (Priest) on Jun 26, 2006 at 13:19 UTC
    I noticed that I had an incorrect superstition about using <I>, namely the brackets, in the left side of s///, as I thought it was normally replaced by reading from a filehandle. It seems to work here, though, so I wonder why I believed that....
[reply]
[d/l]
[select]
      Because Perl is rather context-sensitive and uses <I> to mean that in other contexts? In the code: 
      while (<I>) {
  s/<I>/<italic>/g;
}

      [download]
      the first <I> reads from a filehandle and the second is literal text to replace. Fun, isn't it? :)
[reply]
[d/l]
[select]

Back to Seekers of Perl Wisdom