Regex print first instance 1

[stevio]

Hi there,

Have a simple script which runs a command on the system and returns the regex value

my @val = `command here`;
    
        foreach $line(@val) {           
           if ($line =~ /(test\d+)/){
             print "$1\n";
           }
        }

The question is, I only want to print the first value that it returned if more than one is found, but at the same work through each line. So if the resulting text of the command was:

__DATA__

Test1
Test6
Test5
Test1

I only want to print out the first Test1

Any help appreciated

 

[KevinADC]

Use a hash to store the regexp matches and skip any that is already in the hash.

my @val = `command here`;
my %seen;    
foreach $line(@val) {           
   if ($line =~ /(test\d+)/){
      print "$1\n" if ++$seen{$1} == 1;
   }
}

------------------------------------------
- Kevin, perl coder unexceptional!

 

[stevio]

Thanks Kevin, did the job perfectly

 

[WinblowsME]

if ($line =~ /(test\d+)/[COLOR=red]i[/color]){

 

[WinblowsME]

Or maybe this would work better...

if ( lc ( $line ) =~ /(test\d+)/ )