I have a long meaningless string where every character is either a number [0-9] or a lowercase letter [a-z] as shown below
"0z8b816ne139z1b948bjk50f9498t139gjj90t7tb3509w6h0r7tbp"
I want a regular expression that can match non-consecutive patterns that appear more than once in a string I want the output to look like this
The bold part is the matching part
"0z8b816ne139z1b948bjk50f9498t139gjj90t7tb3509w6h0r7tbp"
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
Regular expression:
(.. )(?=.*?(\1))Reference link
const regex = /(..+)(?=.*?())/gm; // 使用RegExp构造函数的替代语法 // const regex = new RegExp('(..+)(?=.*?(\1))', 'gm') const str = `0z8b816ne139z1b948bjk50f9498t139gjj90t7tb3509w6h0r7tbp `; let m; while ((m = regex.exec(str)) !== null) { // 避免零宽匹配导致无限循环 if (m.index === regex.lastIndex) { regex.lastIndex++; } // 可以通过`m`变量访问结果 m.forEach((match, groupIndex) => { console.log(`找到匹配，第${groupIndex}组：${match}`); }); }