perl use variable under a pattern regex

Question

How to use variable in the pattern ?

Please I want to this :

my $project="SDK" //or something that i will get it after calling some function.
my $JIRA_regex = '(^| )($project)-(\d+)';
print "pattern = $JIRA_regex\n";

Output is not good :

(^| )($project)-(\d+)

Thank you :)

1 - Yes I want to use $project, as string value to match or a regex too:

2 - $JIRA_regex will be matched further on the code.

This is my code that it works fine now :

my $repo=$ARGV[0];
my $comment=$ARGV[1];

my $project_pattern="[A-Z]{2,5}";

if ($repo =~ "test1.git" or $repo =~ "test2.git")
{
    $project_pattern = "\QSDK\E";
}

my $JIRA_regex = "(^| )($project_pattern)-(\\d+)";

if ( $comment =~ /$JIRA_regex/m )
{
    print "matched $2-$3\n";
}
else
{
    print "not matched\n";
}

See stackoverflow.com/questions/15041877/…

Joshua Goldberg
– Joshua Goldberg

2022-07-18 16:13:08 +00:00
Commented Jul 18, 2022 at 16:13 — Joshua Goldberg
– Joshua Goldberg, Commented Jul 18, 2022 at 16:13

melpomene · Accepted Answer · 2016-10-20 17:19:18Z

4

Single quotes don't interpolate variables; double quotes do.

my $project = "SDK"; # or whatever
my $JIRA_regex = "(^| )($project)-(\\d+)";
print "pattern = $JIRA_regex\n";

(Note that I had to escape the backslash to get a literal \d into the string.)

There are some other things to consider:

Is $project supposed to be interpreted as a regex? (Probably not, in which case it should be wrapped in \Q \E or quotemeta().)
Does $JIRA_regex have to be a plain string? If not, it's easier to make it a regex object.

In which case a better solution would be:

my $project = "SDK"; # or whatever
my $JIRA_regex = qr/(^| )(\Q$project\E)-(\d+)/;
print "pattern = $JIRA_regex\n";

answered Oct 20, 2016 at 17:19

melpomene

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