Question

Here is the substring I am working with

[sitetree_link%20id=2]

I need to replace all occurrences of %20 that fall between [ ] with a blank space. But obviously if there are %20s outside of [ ] braces, leave them alone…

I’m just learning regex now but this one seems pretty tough. Anyone got a super smart regex for this?

Thanks 🙂

Answer 1

You can try this

$result = preg_replace('/(\[[^]]*?)(%20)([^]]*?\])/m', '$1 $3', $subject);

Explanation

(          # Match the regular expression below and capture its match into backreference number 1
   \[         # Match the character “[” literally
   [^]]       # Match any character that is NOT a “]”
      *?         # Between zero and unlimited times, as few times as possible, expanding as needed (lazy)
)
(          # Match the regular expression below and capture its match into backreference number 2
   %20        # Match the characters “%20” literally
)
(          # Match the regular expression below and capture its match into backreference number 3
   [^]]       # Match any character that is NOT a “]”
      *?         # Between zero and unlimited times, as few times as possible, expanding as needed (lazy)
   \]         # Match the character “]” literally
)