Replacing string with regexp without using modules

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

I am currently using this code to replace $a and $b variables from a template. However I need to make this more dynamic so I do not need to specify the values any more.

sub prog_replace_string {
    my $replace_this = shift;
    my $with_this  = shift;
    my $string   = shift;

    my $length = length($string);
    my $target = length($replace_this);

    for(my $i=0; $i<$length - $target + 1; $i++) {
        if(substr($string,$i,$target) eq $replace_this) {
            $string = substr($string,0,$i) . $with_this . substr($stri
+ng,$i+$target);
        }
    }
    return $string;
}
[download]

I was wondering how to do replace certain denoted names dynamically in a template file, such as in this context: $var and replacing this with the value of $var. I would think this would be possible by using a reference to the regexp match in a regexp like: $string = s/\$XXXXXXXX/${$1}/; Of course this does not work, I do not know what to replace XXXXXXXX with and the $1 variable would return the entire variable name, however we need it without the $ sign. This must be really easy. I would love if someone could help me out without pointing to the use of a module. Thanks a lot!

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Re: Replacing string with regexp without using modules by pc88mxer (Vicar) on Aug 05, 2008 at 22:51 UTC
For your first question: `sub prog_replace_string { my ($replace_this, $with_this, $string) = @_; $string =~ s{\Q$replace_this\E}{$with_this}g; $string; }` [download] For your second: `sub replace_vars { my ($string, $hash) = @_; $string =~ s{\$(\w+)}{$$hash{$1}}ge; $string; } my %vars = ( a => 1, b => 22, c => 333 ); print replace_vars('a: $a, b: $b, c: $c', \%vars);` [download]	[reply] [d/l] [select]
Re^2: Replacing string with regexp without using modules by cowgirl (Acolyte) on Aug 06, 2008 at 00:37 UTC
Thanks for the great reply. It works well. However when I cut it down to this, it doesn't work: `$string = '$a $b $c'; $hash{a} = "go"; $string =~ s{\$(\w+)}{$$hash{$1}}ge;` [download] Could it be the hash reference you use to the subroutine? \$hash ? Thanks.	[reply] [d/l]
Re^3: Replacing string with regexp without using modules by almut (Canon) on Aug 06, 2008 at 01:58 UTC
Could it be the hash reference ...? Yes, in pc88mxer's code, `$hash` is a hash reference, so it must be dereferenced. In your case, it's a normal hash. So either don't dereference `$hash{a} = "go"; $string =~ s{\$(\w+)}{$hash{$1}}ge;` [download] or create a reference `$hash->{a} = "go"; $string =~ s{\$(\w+)}{$$hash{$1}}ge;` [download] or even use the normal arrow syntax in both the assignment `$hash->{a} = ...`, and when you access it `$hash->{$1}`: `$hash->{a} = "go"; $string =~ s{\$(\w+)}{$hash->{$1}}ge;` [download]	[reply] [d/l] [select]