正则表达式模式可选字符 [英] Regexp pattern Optional character

问题描述

我想匹配像 19740103-0379 或 197401030379 这样的字符串，即破折号是可选的.如何使用正则表达式完成此操作?

I want to match a string like 19740103-0379 or 197401030379, i.e the dash is optional. How do I accomplish this with regexp?

推荐答案

通常你可以使用 -?.或者，您可以使用 -{0,1} 但您应该发现 ? 几乎在任何地方都支持零次或一次出现".

Usually you can just use -?. Alternatively, you can use -{0,1} but you should find that ? for "zero or one occurrences of" is supported just about everywhere.

pax> echo 19740103-0379 | egrep '19740103\-?0379'
19740103-0379

pax> echo 197401030379 | egrep '19740103\-?0379'
197401030379

如果您想在任何地方接受带有任意数量破折号的 12 位数字，您可能需要执行以下操作:

If you want to accept 12 digits with any number of dashes in there anywhere, you might have to do something like:

-*([0-9]-*){12}

基本上是零个或多个破折号后跟 12 次出现(一个数字后跟零个或多个破折号)，并将捕获各种精彩内容，例如:

which is basically zero or more dashes followed by 12 occurrences of (a digit followed by zero or more dashes) and will capture all sorts of wonderful things like:

--3-53453---34-4534---

(当然，如果您的正则表达式引擎支持，您应该使用 \d 而不是 [0-9]).

(of course, you should use \d instead of [0-9] if your regex engine has support for that).

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