regex - 正则表达式 - 查找可以找到给定数字的所有组

Question

我有一个这样的字符串：

:1-2-35:2-3-1:5-6-27456:35-2-11:9-5-6:1-5-2:

我想获取所有包含数字2 的组字符串总是由 3 个数字组成的组，它们之间有一个破折号。

所以我的正则表达式会返回这个：

1 => :1-2-35:
2 => :2-3-1:
3 => :35-2-11:
4 => :1-5-2:

我试过这个没有成功：:\d*2-|-2-|2-\d*:

谢谢你的帮助。

Answer 1

你可以试试这个正则表达式

[^:]*(?<=[-:])2(?=[-:])[^:]*

[^:]表示匹配任何字符，除了:

[^:]*将匹配 0 到多个字符，除了:

2(?=[-:])仅当它后跟-or时才匹配 2:

(?<=[-:])2将匹配 2 只有它前面是-or:

或者

[^:]*\b2\b[^:]*

Answer 2

If the groups will always contain 3 number (and 2 dashes), you can use a regex like this:

:(2-\d+-\d+|\d+-2-\d+|\d+-\d+-2)(?=:)

(Note, it may vary slightly, based on the regex implementation of the language you are using.)

See, also, this short demo in PHP.

Answer 3

You can use this:

(?<=:)(?:2-\d+-\d+|(?:\d+-){1,2}2\b[^:]*)

Answer 4

The following regex should do the job

(?<=:)(?:2-\d+-\d+)|(?:\d+-2-\d+)|(?:\d+-\d+-2)(?=:)

The only limitation is whilst it filters on the : chars they are not included in the match. If you try and include the : chars in the match then sequential matches will fail because the trailing : will already be gone for the beginning of the next match