如何使正则表达式匹配的一部分成为可选的? [英] How do I make part of a regex match optional?

问题描述

这是一个示例字符串:

123456#p654321

目前，我正在使用此匹配将 123456 和 654321 捕获到两个不同的组中:

Currently, I am using this match to capture 123456 and 654321 in to two different groups:

([0-9].*)#p([0-9].*)

但有时，字符串的 #p654321 部分将不存在，因此我只想捕获第一组.我试图通过将 ? 附加到第二组来使第二组可选"，这是可行的，但前提是剩余字符串的末尾有 #p.

But on occasions, the #p654321 part of the string will not be there, so I will only want to capture the first group. I tried to make the second group "optional" by appending ? to it, which works, but only as long as there is a #p at the end of the remaining string.

解决此问题的最佳方法是什么?

What would be the best way to solve this problem?

推荐答案

您在捕获组之外拥有 #p，这使其成为结果的必需部分.您还错误地使用了点字符 (.).点(在大多数 reg-ex 变体中)将匹配任何字符.改为:

You have the #p outside of the capturing group, which makes it a required piece of the result. You are also using the dot character (.) improperly. Dot (in most reg-ex variants) will match any character. Change it to:

([0-9]*)(?:#p([0-9]*))?

(?:) 语法是您获得非捕获组的方式.然后我们只捕获您感兴趣的数字.最后，我们将整个内容设为可选.

The (?:) syntax is how you get a non-capturing group. We then capture just the digits that you're interested in. Finally, we make the whole thing optional.

此外，大多数 reg-ex 变体都有一个用于数字的 d 字符类.所以你可以进一步简化:

Also, most reg-ex variants have a d character class for digits. So you could simplify even further:

(d*)(?:#p(d*))?

正如另一个人指出的那样，* 运算符可能会匹配 零 个数字.为防止出现这种情况，请改用 + 运算符:

As another person has pointed out, the * operator could potentially match zero digits. To prevent this, use the + operator instead:

(d+)(?:#p(d+))?

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