[gmadhukarreddy]

How can I write something like

$a =~ s/\d+/eval($b+$c-$d)/g;

where i need to get the evaluated value to be substituted. but it is not evaluating....

[Chas]

Originally Posted by gmadhukarreddy

How can I write something like

$a =~ s/\d+/eval($b+$c-$d)/g;

where i need to get the evaluated value to be substituted. but it is not evaluating....

I'm not exactly sure of what you are trying to do. For what you have there, I would do something like this:

Code:

my $regex = $b +$c - $d;
$a =~ s/\d+/$regex/g;

Can you not get away with doing this though:

Code:

$a = $b + $c - $d;

??

Maybe I'm missing your point?

These might help:
[regex links]
* http://www.perldoc.com/perl5.8.0/pod...lar-expression
http://www.perldoc.com/perl5.8.0/pod/perlfaq6.html
http://www.perldoc.com/perl5.8.0/pod/perlre.html
http://www.perldoc.com/perl5.8.0/pod/perlrequick.html
http://www.perldoc.com/perl5.8.0/pod/perlretut.html
[/regex links]

* This one in particular.


~Charlie

[gmadhukarreddy]

actually, i have a line like ....

123..345,456..567,678..789

and i have to substitute each number (like 123) by adding something to it.

so if i have to add 20 to each,

the regular expression should substitute 123 by 143 (i.e.123+20 )
345 by 365
456 by 476 and so on

so how can I write the regular expression ????

[Chas]

so how can I write the regular expression ????

Check out the link I posted above with a star on it:

Code:

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

our $val = 0;
my $string = "123..345,456..567,678..789";

print "Content-Type: Text/HTML\n\n";
print "<pre>\n";
print "Original string: $string\n";

$string =~ s/(\d+)(?{$val = $1 + 20;})/$val/g;

print "New string: $string\n";
print "</pre>";

~Charlie

[gmadhukarreddy]

your code works great.......thanks charlie

[Chas]

Originally Posted by gmadhukarreddy

thanks charlie

No worries.

~Charlie