Re: Replacing string with regexp without using modules

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]

Comment on Re: Replacing string with regexp without using modules Select or Download Code

Replies are listed 'Best First'.
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]


good chemistry is complicated, and a little bit messy -LW
	PerlMonks