How do i interpolate a variable in a regular expression?

sanjay nayak has asked for the wisdom of the Perl Monks concerning the following question:

Suppose $method =INVITE.

my %config = (
  URI =>   { regex => '^$method\ssip:', sub => \&request_line },
)
[download]

It doesn't interpolate the variable $method?
Suggest me some code.

Comment on How do i interpolate a variable in a regular expression? Download Code

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Re: How do i interpolate a variable in a regular expression? by davido (Cardinal) on Sep 27, 2006 at 06:52 UTC
Single quotes don't interpolate. Double quotes do. For example, `my $not_interpolated = '$this';` won't interpolate, but `my $interpolated = "$this";` will. But you may find it helpful to use qr// (as discussed in perlop). ...used as follows: `my %config = ( URI => { regex => qr/^$method\ssip:/, sub => \&request_line, }, );` [download] Dave	[reply] [d/l] [select]
Re^2: How do i interpolate a variable in a regular expression? by sanjay nayak (Sexton) on Sep 27, 2006 at 07:04 UTC
Hi Thanks a lot.	[reply]
Re: How do i interpolate a variable in a regular expression? by reneeb (Chaplain) on Sep 27, 2006 at 06:54 UTC
You should use the qr-operator: `my %config = ( URI => { regex => qr/^$method\ssip:/, sub => \&request_line }, )` [download] see perldoc perlop	[reply] [d/l]
Re: How do i interpolate a variable in a regular expression? by chargrill (Parson) on Sep 27, 2006 at 06:54 UTC
For starters, your `$`scalar is in single quotes, which are notorious for not inerpolating variables. --chargrill `s*lil; $=join'',sort split q; s;.;grr; &&s+(.(.)).+$2$1+; $; = qq-$_-;s,.,ahc,;$,.=chop for split q,,,reverse;print for($,,$;,$,$/)` [download]	[reply] [d/l] [select]