PHP preg_replace - Replace text part but not too soon

Question

Using:

$text = preg_replace("/\[\[(.*?)SPACE(.*?)\]\]/im",'$2',$text);

for cleaning and get wordtwo

$text = '..text.. [[wordoneSPACE**wordtwo**]] ..moretext..';

but fails if text has [[ before

$text = '.. [[ ..text(not to cut).. [[wordoneSPACE**wordtwo**]] ..moretext..';

how can I limit to only where I have only the SPACE word?

Answer 1

If there can be no [ and ] inside the [[...]] you may use

$text = preg_replace("/\[\[([^][]*)SPACE([^][]*)]]/i",'$2',$text);

See the regex demo. [^][] negated character class will only match a char other than [ and ] and won't cross the [[...]] border.

Otherwise, use a tempered greedy token:

$text = preg_replace("/\[\[((?:(?!\[{2}).)*?)SPACE(.*?)]]/is",'$2',$text);

See this regex demo.

The (?:(?!\[{2}).)*? pattern will match any char, 0 or more repetitions but as few as possible, that does not start [[ char sequence, and won't cross the next entity [[ border.

Answer 2

Another option might be using possessive quantifiers.

In the first group you could use a negated character class to match any characters except square brackets or an S if it is followed by SPACE.

\[\[([^][S]++(?:S(?!PACE)|[^][S]+)*+)SPACE([^][]++)\]\]

In parts

• \[\[ Match [[
• ( Capture group 1
  • [^][S]++ Match 1+ times any char except S, ] or [
  • (?: Non capturing group
    • S(?!PACE) Match either an S not followed by PACE
    • | Or
    • [^][S]+ Match 1+ times any char except S, ] or [
  • )*+ Close group and repeat 0+ times
• ) Close group 1
• SPACE Match literally
• ( Capture group 2
  • [^][]++ Match 1+ times any char except ] or [
• ) Close group
• \]\] Match ]]

Regex demo