字符串替换正则表达式仅在 [] 之外使用标记

发布于 2024-12-02 06:46:55 字数 212 浏览 1 评论 0原文

我正在尝试用“+”替换字符串中的所有空格。但它应该忽略 [] 内的空格。有谁知道如何使用正则表达式来做到这一点。

示例：

latitude[0 TO *] longitude[10 TO *]

应替换为

latitude[0 TO *]+longitude[10 TO *]

原文

I am trying to replace all the spaces in string with '+'. But it should ignore ignore the spaces inside the []. Do anyone know how to do this using regular expression.

Example:

latitude[0 TO *] longitude[10 TO *]

should be replaced as

latitude[0 TO *]+longitude[10 TO *]

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纵性 2024-12-09 06:46:56

将 ("global" 标志已启用) 替换

\s(?=[^\]]*(?:\[|$))

为

Explanation

\s          # white-space (use an actual space if you want)
(?=         # but only when followed by (i.e. look-ahead)...
  [^\]]*    #   ...anything but a "]"
  (?:       #   non-apturing group ("either of")
    \[      #     "["
    |       #     or
    $       #     the end of the string
  )         #   end non-capturing group
)           # end look-ahead

这会匹配不在方括号内的任何空格，前提是字符串中没有不正确打开/关闭的方括号，并且方括号从不嵌套。

Replace ("global" flag enabled)

\s(?=[^\]]*(?:\[|$))

with

Explanation

\s          # white-space (use an actual space if you want)
(?=         # but only when followed by (i.e. look-ahead)...
  [^\]]*    #   ...anything but a "]"
  (?:       #   non-apturing group ("either of")
    \[      #     "["
    |       #     or
    $       #     the end of the string
  )         #   end non-capturing group
)           # end look-ahead

This matches any space that is not inside a square bracket, under the assumption that there are no incorrectly opened/closed square brackets in the string and that square brackets are never nested.

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