正则表达式仅在字符串未转义时匹配字符串中的字符 [英] RegEx for matching a character in a string only if it isn't escaped

问题描述

如果字符串中的字符没有被转义(即以另一个字符的奇数开头)，我如何才能找到并对其进行操作?

How can I find and operate on a character in a string only if it isn't escaped (ie proceeded by an odd number of another character)?

示例:

所需字符:|

转义符:\

| 应该被找到(并对其进行操作，例如 split)

| should be found (and operated on, for example split)

\| 应该不

\\| 应该

\\\| 应该不

推荐答案

使用否定后视来定义边界:

Use a negative a lookbehind to define a boundary:

(?<!\\)(?:\\\\)*\|

查看现场演示

注意 Java 中的反斜杠，在正则表达式之上是:

Taking care of backslashes in Java, above regex would be:

(?<!\\\\)(?:\\\\\\\\)*\\|

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