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
??
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.
Three changes:
e
modifier
to allow an expression in the
replacement part.g
modifier. This is not needed if you've one date per line.$1
on the replacement side, not a backreferenceThis 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.
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;
$
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With