See perlfaq4: "How do I strip blank space from the beginning/end of a string?".

        s/^\s+//; # strip white space from the beginning
        s/\s+$//; # strip white space from the end
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I just came back to post I found a working solution:

        $buff =~ s/ *$//;  #Remove Trailing Spaces
        $buff =~ s/^ *//;  #Remove Leading Spaces
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You've got the same idea I have, but I'm too regex newbie to figure out the differences. Can you tell me if there are any differences in our regexes?
Thanks!!

/ *$/ means 'match zero or more space characters anchored to the end of the string', which will always match since it can always match zero characters. It will also match only space characters - other white space (such as tabs) will not match.

/\s+$/ means 'match one or more white space characters anchored to the end of the string'. The \s includes spaces, tabs, and other forms of white space.

See perlre for more information. The Tutorials section also has several entries in Pattern Matching, Regular Expressions, and Parsing, which you may find helpful.

You are matching only space characters. bobf's version matches any white space characters (which includes spaces, tabs and various other characters including UNICODE white space characters).

---

DWIM is Perl's answer to Gödel

Perhaps the biggest and most generally used tool in the Perl toolbox is the regex (see also perlretut). I strongly recommend that you take a look at the regular expression documentation then if you are still asking this question come back with the regex substitution you are having trouble with.

use String::Util qw(trim)
my $s = trim('  here are lots of nasty spaces        ');
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In what decate was it good practice introducing a dependency on an external module in order to get the exact same two substitutions hidden in a sub:

#---------------------------------------------------------------------
+---------
# trim
# 

=head1 trim(string)

Returns the string with all leading and trailing whitespace removed.
Trim on undef returns undef.

=cut

sub trim{
    my ($val) = @_;
    
    if (defined $val) {
        $val =~ s|^\s+||s;
        $val =~ s|\s+$||s;
    };
    
    return $val;
}
# 
# trim
#---------------------------------------------------------------------
+---------
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If it really bothers you just add one line to your code and be done with it.

In what decate was it good practice introducing a dependency on an external module in order to get the exact same two substitutions hidden in a sub:

The same decade where abstracting operations behind a meaningful name made programs easier to understand. Compare:

# trim whitespace
$string =~ s/^\s+//;
$string =~ s/\s+$//;
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... to:

trim( $string );

Checkout perlfaq4, you may also want to have a look at perlre and perlretut

The size of the strings may change the way you do it, because capturing a 10Gb string just to remove spaces from the start and end may not be the best approach.

Also, what have you tried so far?

With a regex, of course.

s/^\s+//; # removes leading white space
s/\s+$//; # removes trailing white space
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$my_data_line =~ s/^\s+|\s$+//g;
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perl -p -i.bak -e 's/^\s+|\s$+//g' *.html
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you probably mean:

$my_data_line =~ s/^\s+|\s+$//g;
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