> The fix doesn't work because the regex has to decide which non-greedy ? has "precedence".

strange, it works like I initially thought

  DB<63> print join "|",'abc' =~ /a(.+?)(c)?/
b|c
  DB<64> print join "|",'abd' =~ /a(.+?)(c)?/
b|
[download]

well you seem to have other issues I can't see.

EDIT

Interesting

  DB<84> print join "|",'a<c1<c2' =~ /(.+?)<(c.)?/ # ok
a|c1

  DB<85> print join "|",'a<b1<b2' =~ /(.+?)<(c.)?/ # ok
a|

  DB<86> print join "|",'a<b1<c2' =~ /(.+?)<(c.)?/ # oops
a|
[download]

you might want to use use re 'debug' to parse whats happening.

update

ah now it's clearer

  DB<89> print join "|",'a<b1<c2' =~ /(.+?)(c.)?/
a|
[download]

The non-greedy is finishing as soon as it succeeds, and c. is optional

Ok - your explanation makes sense... Thanks (++).

I'll wait to see of other monks can find a way to do what I need with a single regex.

                All power corrupts, but we need electricity.

> can find a way to do what I need with a single regex.

"single regex" is relative in Perl since you can include Perl-code investigating a sub-match

Does it must be a single regex, or is it just academic curiosity?

edit

you need a positive condition which allows to stop the non-greedy only if you have a match or you reached the end.

  DB<113> print join "|",'a<b1<b2' =~ /(a).+?(?:(c.)|$)/
a|

  DB<114> print join "|",'a<b1<c2' =~ /(a).+?(?:(c.)|$)/
a|c2

  DB<115> print join "|",'a<c1<c2' =~ /(a).+?(?:(c.)|$)/
a|c1
[download]

Try to apply this technique.

HTH

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)
Je suis Charlie!}