java - 正则表达式捕获包含太多内容

Question

我有一个字符串，我想从中捕获冒号之后(包括冒号)直到(不包括)空格或括号的所有内容。

为什么以下正则表达式在字符串匹配中包含括号？:(.*?)[\(\)\s] 或 :(.+?)[\)\s] (非贪婪)不起作用.

示例输入:WHERE t.operator_id = :operatorID AND (t.merchant_id = :merchantID) AND t.readerApplication_id = :readerApplicationID AND t.accountType in :accountTypes

应提取:operatorID、:merchantID、:readerApplicationID、:accountTypes。但我的正则表达式提取了第二场比赛 :marchantID)出了什么问题以及为什么？

即使我在捕获中使用更精确的映射条件，它也不起作用:
:([a-zA-z0-9_]+?)[\)\(\s]

Answer 1

将条件“后跟空格或括号”设置为 lookahead ，这样它可以看到但不匹配。现在，您正在将括号与 [\(\)\s] 显式匹配:

:(.+?)(?=[\s\(\)])

https://regex101.com/r/im8KWF/1/

或者，使用内置的 \b “word boundary ”，这也是一个“零宽度”断言，含义相同*:

:(.+?)\b

https://regex101.com/r/FnnzGM/3/

*正则表达式.info 中的字边界定义:

There are three different positions that qualify as word boundaries:

Before the first character in the string, if the first character is a word character. After the last character in the string, if the last character is a word character. Between two characters in the string, where one is a word character and the other is not a word character.