regexp $1 in a variable

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

Hi Monks,

still "only" a regexp question:

I have

$a='abab';
$b='a(.)';
$c='a$1';

print "\$a $a\n";
print "\$b $b\n";
print "\$c $c\n";

$a =~ s/$b/$c/g ;    # so this is  s/a(.)/a$1/g

print "\$a $a\n";
[download]

which prints

$a abab
$b a(.)
$c a$1
$a a$1a$1
[download]

What I wanted/expected was the last line to be

$a abab
[download]

It's clear, after $c is evaluated to a$1 $1 isn't further evaluated to the content of the matching braces in the regexp.

How to change the above code so that the output is my expected one? Is it possible to combine this two different evaluations? Thanks for any help.

regards,
almaric

Comment on regexp $1 in a variable Select or Download Code

Replies are listed 'Best First'.
Re: regexp $1 in a variable by Enlil (Parson) on Mar 02, 2004 at 03:51 UTC
You might find $1 in variable regex replacement string of interest for your particular problem. -enlil	[reply]
Re: regexp $1 in a variable by Stevie-O (Friar) on Mar 02, 2004 at 02:05 UTC
`$a =~ s/$b/qq[qq{$c}]/eeg` [download] Not sure exactly why the double-e (and double-qq) are needed, but what can I say -- it works! --Stevie-O `$"=$,,$_=q>\|\p4<6 8p<M/_\|<('=> .q>.<4-KI<l\|2$<6%s!<qn#F<>;$, .=pack'N',"@{[unpack'C',$_] }"for split/</;$_=$,,y[A-Z a-z] {}cd;print lc` [download]	[reply] [d/l] [select]
Re: Re: regexp $1 in a variable by mawe (Hermit) on Mar 02, 2004 at 07:14 UTC
Maybe this looks a bit cleaner :-) `... $c='"a$1"'; ... $a =~ s/$b/$c/eeg ; ...` [download]	[reply] [d/l]
Re: regexp $1 in a variable by Roger (Parson) on Mar 02, 2004 at 00:36 UTC
~~`$c=qr/a$1/;`~~ `eval "$a =~ s/$b/$c/g";` [download]	[reply] [d/l] [select]
Re: regexp $1 in a variable by esskar (Deacon) on Mar 02, 2004 at 00:29 UTC
change `$b='\Qa(.)\E';` [download] and it will work UPDATE: wrong thought	[reply] [d/l]