正则表达式查找包含在两个字符之间的字符串，同时排除分隔符 [英] Regular Expression to find a string included between two characters while EXCLUDING the delimiters

问题描述

我需要从一个字符串中提取一组包含在两个分隔符之间的字符，而不返回分隔符本身.

I need to extract from a string a set of characters which are included between two delimiters, without returning the delimiters themselves.

一个简单的例子应该会有所帮助:

A simple example should be helpful:

目标:提取方括号之间的子串，不返回方括号本身.

Target: extract the substring between square brackets, without returning the brackets themselves.

基本字符串:这是一个测试字符串[或多或少]

如果我使用以下 reg.例如

If I use the following reg. ex.

[.*?]

匹配是[或多或少].我只需要获得 more or less (没有括号).

The match is [more or less]. I need to get only more or less (without the brackets).

可以吗?

推荐答案

简单易行:

(?<=[)(.*?)(?=])

从技术上讲，这是使用前瞻和后视.请参阅前瞻和后视零宽度断言.该模式包括:

Technically that's using lookaheads and lookbehinds. See Lookahead and Lookbehind Zero-Width Assertions. The pattern consists of:

• 前面有一个 [ 未被捕获(后视)；
• 一个非贪婪的捕获组.在第一个停止是非贪婪的];和
• 后跟一个未被捕获的 ](前瞻).

或者，您可以只捕获方括号之间的内容:

Alternatively you can just capture what's between the square brackets:

[(.*?)]

并返回第一个捕获的组而不是整个匹配.

and return the first captured group instead of the entire match.

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