Regx

[jain236]

Hi all

i am facing problem in the regex when is substitute a variable in the text.

here is the code

$ver = 1.92 (i get it dynamically)
 my $patternph = 'Configuration file '. "\"phone1\.cfg\"" ." is from template phone1\.cfg, revision "."$phver" ;
foreach my $val (@output)
    {    
 
        if ($val =~ /\Q$patternph\E/)
        {
            $phcount++;
        }elsif($val =~ /\Q$patternsip\E/)
        {
            $sicount++;                
        }
 
    }
 

@output contains lot of data the in which i am looking is as follows

 69|0723111807|cfg  |*|02|Prm|Configuration file "phone1.cfg" is from template phone1.cfg, revision 1.92
 
 

i am not why it is failing, if i dont append the $phver in the last then it is finding.

can some one help in this regards?

[numberwhun]

Have you tried removing the double quotes from around $phver and see if that works?

I would make sure that the variable $patternph is getting set right by testing it with print statements so you can see it.

Regards,

Jeff

[nithinpes]

You have assigned the version number to $ver variable, why have you used $phver in your pattern?
Also, if you are getting $ver dynamically from file or through STDIN, do remember to strip newline using chomp.
Apart from that, I don't see anything wrong in th regex.

[KevinADC]

side note, use quoting operators to make your perl code more readable, instead of this monstrosity of a construct:

my $patternph = 'Configuration file '. "\"phone1\.cfg\"" ." is from template phone1\.cfg, revision "."$phver" ;

You can do this:

my $patternph = qq{Configuration file "phone1.cfg" is from template phone1.cfg, revision $ver} ;

perldoc: Quote Like Operators