Perl Remove Portion of String

This appears to work for me:

#!/usr/bin/perl

use strict;
use warnings;

my $variable = "{MD5}lL5Bo6QZUIcsEuxusMBXMR1YAgCAZF== {MD5}BY2I75mxmgp1DPwsTYKiargac7==";
print "Variable before: $variable\n";
$variable =~ s/\s\{MD5\}.*//;
print "Variable after: $variable\n";

OUTPUT:
Variable before: {MD5}lL5Bo6QZUIcsEuxusMBXMR1YAgCAZF== {MD5}BY2I75mxmgp1DPwsTYKiargac7==
Variable after: {MD5}lL5Bo6QZUIcsEuxusMBXMR1YAgCAZF==

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Will the fields always end with == or always start with {MD5}?

Here's code that will work for either (with comments).

#!/usr/local/bin/perl

# always use strict and warnings
use strict;
use warnings;

my $variable = '{MD5}lL5Bo6QZUIcsEuxusMBXMR1YAgCAZF== {MD5}BY2I75mxmgp1DPwsTYKiargac7==';

# you probably want to comment out all of these except one

# if always ending in ==
$variable =~ s{(?<==).*$}{};

# if always starts with {MD5} - make sure it's not the first one
# I couldn't figure out how to get a look-behind to match any char
$variable =~ s{(.){MD5}.*$}{$1};

# if there is always a space between items
$variable =~ s{\s.*$}{};

print $variable, "\n";

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All options seem to work equally as well, but I'm focusing on

$variable =~ s{(.){MD5}.*$}{$1};

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since I think it will always start with {MD5}.  I'm confused on how this works and would appreciate a detailed explanation.  Heres my take, although most likely inaccurate:

1. The brackets (except those around MD5) are just the delimenters you chose to use instead of the standard /.  Not sure why the backets around MD5 do not need to be escaped, but it appears they does not.

2. The

(.){MD5}.*${$1}

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section has me confused.  I'm just not sure why this is working.  In the brackets that have $1, I can put {$1} or {} and still get the desired result.  I don't understand how this is removing the second field.

Also, I can remove the $, which I assume is a end of line anchor and still acheive the desired results.  I just want to make sure that it will always remove the second field.

This regx works even if the space is removed between the two feilds, which if great, but I don't see how.  The explanation that I have found on what (.) means doesn't seem to match how this is working.  Thanks

OK.  Great explanation.  Now that I understand what this does I have added the following to make sure this behaves the same way if by chance $variable has a space in the front for some reason.

sub trim {
    $_[0]=~s/^\s+//;
    $_[0]=~s/\s+$//;
    $_[0]=~s/\s+//g;
    return;
.....
}
elsif ($variable =~ /\{MD5\}BY2I75mxmgp1DPwsTYKiargac7==/i {
      print "Variable is $variable\n;
      trim ($variable);
      $variable=~ s{(.){MD5}.*$}{$1};
      print "Variable is $variable\n;
}

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This seems to work pretty well.  It removes any possible leading space so I can be assured the regular expression is not grabbing the wrong field. Also it gets rid of any space in the middle which I do not need and I'm not sure if there could be none, 1 or several spaces.  I don't think I'm missing anything?

$_[0]=~s/^\s+//; # removes leading spaces