Create Custom Regex

[BrianAtWork]

Good morning,

I am writing a perl script that is a wrapper for another Unix command. I want to pass my perl script a regex-style string, and then have the perl script modify that string to pass to the Unix command.

The Unix command uses the percent sign (%) as a wildcard instead of the asterisk (*).

For example:

If I want the Unix command to search for the string "string" or "this%string", then I want my perl script to modify the strings like this: "%string%" or "%this%string%".

Or, if I pass the script "^string" - I want the script to modify it as "string%".

And, "string$" would become "%string"

And, "^string$" would become simply "string".

Now, to accomplish this, I could go about it like this:

if ($string !~/^\^/) { #doesn't have a ^ at the beginning
  $string="\%$string"; #add the leading %
}
else {
  $string=~s/^\^//; #remove the leading ^
}
if ($string !~/\$$/) { #doesn't have a $ at the end
  $string="$string\%"; #add the trailing %
}
else {
  $string=~s/\$$//; #remove the trailing $
}

I think that looks tacky, and I think there has to be a better way to do it...

There's probably a way to do what I want in 2 simple regular expressions, but I'm not seeing the logic.

Does anyone have an idea?

Thanks in advance!

Brian

Oh, and I'm doing this line as well, to turn * into %.

string=~s/\*/\%/g;

That way users can use regex style wildcards, and they will automatically be converted to the wildcard the unix command uses.