Perl regexp match word in string

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hi i try to match expression but not match

What I have tried:

<pre>#!/usr/bin/perl
use warnings;
use strict;
my $s = "Regular expression";
$s =~ /Regular(.*?)/;
print $s;

Posted 22-Jul-22 4:11am

armanvaneld

Updated 22-Jul-22 6:38am

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Patrice T · Answer 1 · 2022-07-22T04:20:00

Quote:
hi i try to match expression but not match

It works :
!
Try

PERL

$s =~ /Regular(.*)/;

Pay attention to the '?' removed, it may give better results.

Just a few interesting links to help building and debugging RegEx.
Here is a link to RegEx documentation:
perlre - perldoc.perl.org[^]
Here is links to tools to help build RegEx and debug them:
.NET Regex Tester - Regex Storm[^]
Expresso Regular Expression Tool[^]
RegExr: Learn, Build, & Test RegEx[^]
Online regex tester and debugger: PHP, PCRE, Python, Golang and JavaScript[^]
This one show you the RegEx as a nice graph which is really helpful to understand what is doing a RegEx:
Debuggex: Online visual regex tester. JavaScript, Python, and PCRE.[^]
This site also show the Regex in a nice graph but can't test what match the RegEx:
Regexper[^]

Richard MacCutchan · Answer 2 · 2022-07-22T04:58:00

Solution 2

That works fine for me, so I cannot see what you may be doing wrong. You also need to understand that the expression $s =~ /Regular(.*?)/ returns 1 for a match, and nothing for a non-match.
Match:

PERL

my $s = "Regular expression";
print $s =~ /Regular(.*?)/ . "\n";

// output:
$ perl perltest.pl
1
$

Not match:

PERL

my $s = "Dogular expression";
print $s =~ /Regular(.*?)/ . "\n";

// output:
$ perl perltest.pl
 
$

Posted 22-Jul-22 4:58am

Richard MacCutchan

Updated 22-Jul-22 4:59am

v2

Comments

[no name] 22-Jul-22 11:08am

thanks for rply but if i want return not 1 but the word espression ? how is possible to do that ?

Richard MacCutchan 22-Jul-22 11:22am

You get what the =~ operator provides. If you want to capture data then see perlre - Perl regular expressions - Perldoc Browser[^].

[no name] 22-Jul-22 12:29pm

i am novice where i find it ??

Richard MacCutchan 22-Jul-22 12:30pm

I don't know, I have not read the document. Google may find you some simpler examples.

k5054 · Answer 3 · 2022-07-22T06:38:00

Solution 3

To assign matched substrings you have to use the substitute regex operator

PERL

my $s = "Regular expression";
$s =~ s/Regular(.*?)/$1/; # note leading s for regex operation
# update: (oops!)  changed assignment (=) to match-assign (=~) in above statement
#update 2:removed spurious $ from match expression
#        was previously /$Regular(.*?)/$1/ Wrong!
print "s = '$s'\n";

Note that $s includes the the whitespace following the matched text, which may not be what you want. But this example should get you going. If you don't want to change the value of $s then the following might be useful

PERL

$s = "Regular expression";
($t = $s) =~ s/Regular(.*?)/$1/;

print "s = '$s'\n";
print "t = '$t'\n";

produces

s = 'Regular expression'
t = ' expression'

Posted 22-Jul-22 6:38am

k5054

Updated 23-Jul-22 1:01am

v3

Comments

Richard MacCutchan 23-Jul-22 4:02am

When I try the code in your first example I get:

Global symbol "$Regular" requires explicit package name (did you forget to declare "my $Regular"?) at perltest.pl line 5.
Execution of perltest.pl aborted due to compilation errors.

Any idea what I am doing wrong?

k5054 23-Jul-22 6:57am

Nothing. That $ before Regular is a typo too. Of course, it works fine if you you don't have use strict; in effect.

Richard MacCutchan 23-Jul-22 14:10pm

Thanks. I assumed it was there to indicate start searching for a match at the beginning of the string.

[no name] 25-Jul-22 3:31am

thanks for rply, anoter question suppose you have a phrases like this Regular expression bao bab lapos you want macth only expression why if i modify the regexp in this mode not work ? $s =~ s/Regular(.*?) b/$1/; thanks

Perl regexp match word in string

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