find and replace text in a perl script

Question

so I know that there are many ways of doing what I am asking for, but all that I have found is not helping me for what I am trying to do. It is supposed to be a simple find and replace script using stdin and stdout. I have a script called replace.pl and this is what i have in it:

#!/usr/bin/perl
use strict;
use warnings;

while(<STDIN>){
  $_ = s/$ARGV[0]/$ARGV[1]/g;
  print STDOUT $_;
}

When I run echo "replace a with b please" | replace.pl 'a' 'b' all I get back is a "1". My desire output is "replace b with b please" but what ever I try to do, it is not changing it. Could any one tell me what i am doing wrong here?

mpapec · Accepted Answer · 2014-05-24 06:45:58Z

3

Try

s/$ARGV[0]/$ARGV[1]/g;

instead of

$_ = s/$ARGV[0]/$ARGV[1]/g;

as s/// returns 1 when substitution was successful.

You can also quote search pattern if it should be literal string (not a regex),

  s/\Q$ARGV[0]\E/$ARGV[1]/g;

edited May 24, 2014 at 6:45

answered May 24, 2014 at 6:37

mpapec

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1 Comment

Doctor06 Over a year ago

oh my gosh... lol thanks. I almost freaking had it. I cant believe i missed that. thank you.

Alex Trebek · Accepted Answer · 2014-05-24 06:49:31Z

1

From perlop:

s/PATTERN/REPLACEMENT/msixpodualgcer

Searches a string for a pattern, and if found, replaces that pattern with the replacement text and returns the number of substitutions made. Otherwise it returns false (specifically, the empty string).

That's why the code sets $_ = 1. You just want to do s/$ARGV[0]/$ARGV[1]/g; for its substitution side-effect, without assigning its return value to $_.

while (<STDIN>) {
    s/$ARGV[0]/$ARGV[1]/g;
    print;
}

edited May 24, 2014 at 6:49

answered May 24, 2014 at 6:44

Alex Trebek

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