用于跳过捕获组中字符的正则表达式 [英] Regular expression to skip character in capture group

问题描述

是否可以在正则表达式中跳过捕获组中的几个字符?我正在使用 .NET 正则表达式，但这无关紧要.

Is it possible to skip a couple of characters in a capture group in regular expressions? I am using .NET regexes but that shouldn't matter.

基本上，我正在寻找的是:

Basically, what I am looking for is:

[随机文字]AB-123[随机文字]

[random text]AB-123[random text]

我需要捕获AB123"，不带连字符.

and I need to capture 'AB123', without the hyphen.

我知道 AB 是 2 或 3 个大写字符，而 123 是 2 或 3 位数字，但这不是难点.困难的部分(至少对我而言)是跳过连字符.

I know that AB is 2 or 3 uppercase characters and 123 is 2 or 3 digits, but that's not the hard part. The hard part (at least for me) is skipping the hyphen.

我想我可以分别捕获两者，然后在代码中将它们连接起来，但我希望我有一个更优雅的、仅使用正则表达式的解决方案.

I guess I could capture both separately and then concatenate them in code, but I wish I had a more elegant, regex-only solution.

有什么建议吗?

推荐答案

简而言之:你不能.匹配始终是连续的，即使它包含零宽度断言等内容，如果您想找到下一个字符，则无法匹配下一个字符.

In short: You can't. A match is always consecutive, even when it contains things as zero-width assertions there is no way around matching the next character if you want to get to the one after it.

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