正则表达式:括号中的必需字符 [英] regex: required character in brackets

问题描述

如何从包括 x 的集合中搜索仅包含字符的字符串，并要求它包含 x?例如[a-z]+ 但如果不包含 x 则不匹配.

How do I search for a string that contains only chars from a set including x, and require it to contain x? e.g. [a-z]+ but not matching if it doesn't contain x.

所以它应该匹配 quux 而不是 foo 或 bar.

So it should match quux but not foo or bar.

推荐答案

假设你想匹配一个只有 x,y 或 z 的字符串，匹配 start (^) 后跟零个或多个 y 或 z ([yz]*) 后跟一个 x 后跟零个或多个 x、y 或 z ([xyz]*) 后跟 end ($)

Assuming you want to match a string of only x,y or z, match start (^) followed by zero or more y or z ([yz]*) followed by an x followed by zero or more x, y or z ([xyz]*) followed by end ($)

^[yz]*x[xyz]*$

如果您尝试匹配 [a-z] 但在某处带有 x，那么应该这样做

If you are trying to match [a-z] but with an x in there somewhere, then this should do it

^[a-z]*x[a-z]*$

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