将不包含特定文本的行与正则表达式匹配 [英] matching a line that doesn't contain specific text with regular expressions

问题描述

我想使用正则表达式执行以下操作，但不确定如何执行.当 one two 是行的开头时，我希望它匹配 one two 除非字符串在 one two<之后的任何位置包含 three/代码>.

I want to do the following with regular expressions but not sure how to do it. I want it to match one two when one two is the beginning of the line unless the string contains three anywhere after one two.

推荐答案

您需要一个 否定前瞻断言 - 类似这样:

You need a negative lookahead assertion - something like this:

/^one two(?!.*three)/m

这是关于前瞻/后视断言的教程

注意:我添加了 'm' 修饰符，以便 ^ 匹配一行的开头而不是整个字符串的开头.

Note: I've added the 'm' modifier so that ^ matches the start of a line rather than the start of the whole string.

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