How do I substitute with an evaluated expression in Perl?

问题

There's a file dummy.txt

The contents are:

 9/0/2010
 9/2/2010
 10/11/2010

I have to change the month portion (0,2,11) to +1, ie, (1,3,12) I wrote the substitution regex as follows

 $line =~ s/\/(\d+)\//\/\1+1\//;

It's is printing

9/0+1/2010
9/2+1/2010
10/11+1/2010

How to make it add - 3 numerically than perform string concat? 2+1??

回答1:

Three changes:

You'll have to use the e modifier to allow an expression in the replacement part.
To make the replacement globally you should use the g modifier. This is not needed if you've one date per line.
You use $1 on the replacement side, not a backreference

This should work:

$line =~ s{/(\d+)/}{'/'.($1+1).'/'}eg;

Also if your regex contains the delimiter you're using(/ in your case), it's better to choose a different delimiter ({} above), this way you don't have to escape the delimiter in the regex making your regex clean.

回答2:

How about this?

$ cat date.txt 
9/0/2010
9/2/2010
10/11/2010
$ perl chdate.pl 
9/1/2010
9/3/2010
10/12/2010
$ cat chdate.pl 
use strict;
use warnings;

open my $fp, '<', "date.txt" or die $!;

while (<$fp>) {
    chomp;
    my @arr = split (/\//, $_);
    my $temp = $arr[1]+1;
    print "$arr[0]/$temp/$arr[2]\n";
}

close $fp;
$

回答3:

this works: (e is to evaluate the replacement string: see the perlrequick documentation).

$line = '8/10/2010';
$line =~ s!/(\d+)/!('/'.($1+1).'/')!e;

print $line;

It helps to use ! or some other character as the delimiter if your regular expression has / itself.

You can also use, from this question in Can Perl string interpolation perform any expression evaluation?

$line = '8/10/2010';
$line =~ s!/(\d+)/!("/@{[$1+1]}/")!e;

print $line;

but if this is a homework question, be ready to explain when the teacher asks you how you reach this solution.

来源：https://stackoverflow.com/questions/3939788/how-do-i-substitute-with-an-evaluated-expression-in-perl

标签

regex

perl

substitution