I would like to match the following: Match A, optionally followed by any number of characters not containing Z, followed by Z, with the additional restriction that *if* there are characters between A and Z, the character immediately following A must be a space. In case of a match, grab what is between A and Z.

Examples for matching strings:

Example for non-matching strings: The obvious solution, as far I can see, goes like this: 
if(/^A
     (?:
       Z
       |(\s.*?)
     )
     Z$/x) {
  $grabbed=$1//''
}

[download]
I don't like the repeating of Z in this pattern. Any suggestion for a more elegant way to do this?

Update: I just came up with this: 
if(/^A
     (
       \s
       (?:.*?)
     )?
     Z$/x) {
  $grabbed=$1//''
}

[download]
This seems to fulfil the condition. Still, if you have nice alternatives, I would like to see them. Plus, is there a way to write the regexp so that $1 is always defined in case of a match? Right now it is undef if the string to match is AZ (that's why I have to use the // operator).
-- 
Ronald Fischer <ynnor@mm.st>

In reply to This regexp made simpler by rovf

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