正则表达式在包含精确子字符串的括号之间查找整个子字符串 [英] Regex find whole substring between parenthesis containing exact substring

问题描述

例如我有字符串:

"one two  (78-45ack sack); now (87 back sack) follow dollow (59 uhhaaa)"

我只需要带括号的整个子字符串，包含单词"back"，对于这个字符串，它将是:

and I need only whole substring with parenthesis, containing word "back", for this string it will be:

"(87 back sack)"

我试过了:

(\(.*?back.*?\))

但它返回 "(78-45ack sack); now (87 back sack)"

我的正则表达式应该是什么样的?据我所知，它发生的原因是搜索从字符串的开头开始，通常 - how to perform regex to "search" 从字符串的中间，从单词 "back"?

How should my regex look like? As I understood it's happening cause search going from begin of the string, in common - how to perform regex to "search" from the middle of string, from the word "back"?

推荐答案

你可以使用这个基于否定字符类的正则表达式:

You can use this regex based on negated character class:

\([^()]*?back[^()]*\)

• [^()]* 匹配 0 个或多个不是 ( 和 ) 的任何字符，从而确保我们不't 匹配 (...) 对.
• [^()]* matches 0 or more of any character that is not ( and ), thus making sure we don't match across the pair of (...).

RegEx 演示 1

或者，您也可以使用这个基于负正则表达式的正则表达式:

Alternatively, you can also use this regex based on negative regex:

\((?:(?!back|[()]).)*?back[^)]*\)

RegEx 演示 2

• (?!back|[()]) 是否定前瞻，断言当前位置没有 back 或 ( 或) 在文本前面.
• (?:(?!back|\)).)*? 匹配 0 个或多个没有 back 或 ( 或 ) 前面.
• [^)]* 匹配除 ) 之外的任何内容.
• (?!back|[()]) is negative lookahead that asserts that current position doesn't have back or ( or ) ahead in the text.
• (?:(?!back|\)).)*? matches 0 or more of any character that doesn't have back or ( or ) ahead.
• [^)]* matches anything but ).

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