用php正则表达式替换方括号之间的子字符串

这是我正在使用的子字符串

[sitetree_link%20id=2]

我需要将所有出现在[]之间的%20替换为空白。但很明显，如果[]大括号外有%20个，请不要管它们

我现在只是在学习正则表达式，但这个似乎很难。有人为此买了超级聪明的正则表达式吗

谢谢：）

你可以试试这个

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

解释

(          # 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
)

---

啊，差不多了！只需要用一个空的空间来代替，即“代替”“非常感谢：）太好了，也为解释干杯。祝你周末愉快！非常感谢您的解释，这非常有帮助！！