如何在两个以上的另一个字符串之间找到字符串？

Question

我有正则表达式：

let re = /\[(.*?)\] \[(.*?)\] \[(.*?)\] (.*)/gm;
let regex = new RegExp(re);

我的字符串：

[Max] [2017-12-12 15:59 (UTC +02:00)] [Technical issues] Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. 
[Corben Dallas] [2017-12-20 12:48 (UTC +02:00)] [Technical issues] Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. 
[Max] [2017-12-12 15:59 (UTC +02:00)] [Technical issues] Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. 

我需要翻阅每个人的文字。但我的问题是，文本可能包含换行符\n，而我的正则表达式仅将文本复制到行尾。请看这里的例子https://regex101.com/r/t7zV1U/1

Answer 1

码

See regex in use here：切换到PCRE风格以查看字符串中的实际匹配。

\[([^\]]*)\] \[([^\]]*)\] \[([^\]]*)\] ((?:(?!^\[)[\s\S])*)

Usage

var s = `[Max] [2017-12-12 15:59 (UTC +02:00)] [Technical issues] Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. 
[Corben Dallas] [2017-12-20 12:48 (UTC +02:00)] [Technical issues] Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. 
[Max] [2017-12-12 15:59 (UTC +02:00)] [Technical issues] Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. `

var r = /\[([^\]]*)\] \[([^\]]*)\] \[([^\]]*)\] ((?:(?!^\[)[\s\S])*)/gm
let m;

while ((m = r.exec(s)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === r.lastIndex) {
        r.lastIndex++;
    }
    
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

---

说明

首先要注意的是，我将你所有的\[(.*?)\]改为\[([^\]]*)\]。这是因为在这种情况下\[([^\]]*)\]实际上比懒惰量词更好（[^\]]*意味着匹配任何不是]的东西）。

• ((?:(?!^\[)[\s\S])*)将以下内容捕获到捕获组4中 (?:(?!^\[)[\s\S])*匹配以下任意次数（这是一个tempered greedy token） (?!^\[)负向前瞻确保后面的内容不匹配 ^在线的开头断言位置 \[字面上匹配左方括号[ [\s\S]匹配任何字符（包括换行符）

Answer 2

它应该是本RegExp的第1组：\[(.*?)\] \[(.*?)\] \[(.*?)\]/gm

见：https://regex101.com/r/tC6mDp/1

你只需要匹配括号：

• 第1组将是名称
• 第2组的日期
• 第3组的主题是什么？