Pattern Matching in perl

Question

I want to parse some information from the file.

Information in the file:

Rita_bike_house_Sha9

Rita_bike_house

I want to have output like dis

$a = Rita_bike_house and $b = Sha9,

$a = Rita_bike_house and $b = "original"

In order to get that I have used the below code:

$name = @_; # This @_ has all the information from the file that I have shown above. 

#For matching pattern Rita_bike_house_Sha9 
($a, $b) = $name =~  /\w\d+/; 

if ($a ne "" and $b ne "" ) { return ($a,$b) } 
# this statement doesnot work at all as its first condition 
# before the end is not satisified.

Is there any way where I can store "Rita_bike_house" in $a and "Sha9" in $b? I think my regexp is missing with something. Can you suggest anything?

`$name = @_` is a code smell. You probably mean [`($name)=@_`](http://stackoverflow.com/q/10031455/168657). — mob, Jul 08 '13 at 19:23
The `\w` matches `_` (underscore) too, so you need more precise matching rule. — clt60, Jul 08 '13 at 19:25
I tired having pattern match as /\w_\w_\w_([\w\d]+)/. But, it doesnt work. Any suggestions in the pattern match? Thanks — user2498830, Jul 08 '13 at 19:28
`\w` only matches one letter. You need `\w+` to match one or more letters (a word). But `\w` also matches `_` so you're probably better off using `[a-z]` if that is what you mean. — Adrian Pronk, Jul 08 '13 at 19:33

score 2 · Answer 1 · answered Jul 08 '13 at 21:22

Please don't use the variables $a and $b in your code. There are used by sort and will confuse you.

Try:

while( my $line = <DATA> ){
  chomp $line;

  if( $line =~ m{ \A ( \w+ ) _ ( [^_]* \d [^_]* ) \z }msx ){
    my $first = $1;
    my $second = $2;
    print "\$a = $first and \$b = $second\n";
  }else{
    print "\$a = $line and \$b = \"original\"\n";
  }
}

__DATA__
Rita_bike_house_Sha9
Rita_bike_house

clt60 · Answer 2 · 2013-07-08T21:26:49.350

0

Not very nice, but the next:

use strict;
use warnings;

while(<DATA>) {
    chomp;
    next if /^\s*$/;
    my @parts = split(/_/);
    my $b = pop @parts if $parts[$#parts] =~ /\d/;
    $b //= '"original"';
    my $a = join('_', @parts);
    print "\$a = $a and \$b = $b,\n";
}

__DATA__
Rita_bike_house_Sha9
Rita_bike_house

prints:

$a = Rita_bike_house and $b = Sha9,
$a = Rita_bike_house and $b = "original",

edited Jul 08 '13 at 21:26

answered Jul 08 '13 at 19:48

clt60

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Thanks for the input but my I get an error saying terminated at "my $b = pop @parts if $parts[$#parts] =~ /\d/;" I cant use use 5.014; – user2498830 Jul 08 '13 at 20:53
Any suggestion why does it not parse the line "my $b = pop @parts if $parts[$#parts] =~ /\d/;" – user2498830 Jul 08 '13 at 21:00

Abhishek Kulkarni · Answer 3 · 2013-07-09T10:30:32.487

If you are sure that the pattern which is required will always be similar to 'Sha9' and also it will appear at the end then just do a greedy matching....

open FILE, "filename.txt" or die $!;
my @data = <FILE>;
close(<FILE>);
#my $line = "Rita_bike_house_Sha9";
foreach $line (@data)
{
    chomp($line);
    if ($line =~ m/(.*?)(_([a-zA-Z]+[0-9]+))?$/)
    {
        $a = $1;
        $b = $3 ? $3 : "original";
    }
}

Pattern Matching in perl

3 Answers3