Perl Text processing on a variable before its usage - perl

Question

I wrote a perl script whihc will output a list containing similar entries like below:
$var = ' whatever'
$var contains: a single quote, a space, the word whatever, single quote
actually, this is key of a hash and i want to pull the value for the same. but due to the single quotes and a space in betweene, i am not able to pull the hash key value.
So, i want to strip $var as below:
$var = whatever
meaning remove the single quote, the space and the trailing single quote.
so that I can use $var as hash key to pull the respective value.
could you guide me on a perl oneliner for the same.
thnaks.

Answer 1

Here is several ways to do it, but beware - modifying the keys in a hash can end with unwanted results, like:
use strict;
use warnings;
use Data::Dumper;
my $src = {
"a a" => 1,
" a a " => 2,
"' a a '" => 3,
};
print "src: ", Dumper($src);
my $trg;
#$trg{ map { s/^[\s']*(.*?)[\s']*$/$1/; $_ } keys %$src } = values %$src;
print "copy: ", Dumper($trg);
will produce:
src: $VAR1 = {
' a a ' => 2,
'\' a a \'' => 3,
'a a' => 1
};
copy: $VAR1 = {
'a a' => 1
};
Any regex is possible do explain with YAPE::Regex::Explain module. (from CPAN). For the above regex:
use YAPE::Regex::Explain;
print YAPE::Regex::Explain->new( qr(^[\s']*(.*?)[\s']*$) )->explain;
will produce:
The regular expression:
(?-imsx:^[\s']*(.*?)[\s']*$)
matches as follows:
NODE EXPLANATION
----------------------------------------------------------------------
(?-imsx: group, but do not capture (case-sensitive)
(with ^ and $ matching normally) (with . not
matching \n) (matching whitespace and #
normally):
----------------------------------------------------------------------
^ the beginning of the string
----------------------------------------------------------------------
[\s']* any character of: whitespace (\n, \r, \t,
\f, and " "), ''' (0 or more times
(matching the most amount possible))
----------------------------------------------------------------------
( group and capture to \1:
----------------------------------------------------------------------
.*? any character except \n (0 or more times
(matching the least amount possible))
----------------------------------------------------------------------
) end of \1
----------------------------------------------------------------------
[\s']* any character of: whitespace (\n, \r, \t,
\f, and " "), ''' (0 or more times
(matching the most amount possible))
----------------------------------------------------------------------
$ before an optional \n, and the end of the
string
----------------------------------------------------------------------
) end of grouping
----------------------------------------------------------------------
In short the: s/^[\s']*(.*?)[\s']*$/$1/; mean:
at the beginning of the string match whitespaces or apostrophe as much times is possible,
then match anything
match at the end of string whitespaces or apostrophes as much times as possible
and keep the only the "anything" part

Answer 2

#!/usr/bin/perl
$string = "' my string'";
print $string . "\n";
$string =~ s/'//g;
$string =~ s/^ //g;
print $string;
Output
' my string'
my string

Answer 3

$var =~ tr/ '//d;
see: tr operator
or, by regex
$var =~ s/(?:^['\s]+)|'//g;
The latter will keep the spaces in the middle of the word, the former removes all spaces and single quotes.
A short test:
...
$var = q{' what ever'};
$var =~ s/
(?: # find the following group
^ # at string begin, followed by
['\s]+ # space or single quote, one or more
) # close group
| # OR
' # single quotes in the while string
//gx ; # replace by nothing, use formatted regex (x)
print "|$var|\n";
...
prints:
|what ever|
as expected.