javascript regexp：匹配 :: 之后的所有内容，但不匹配 >

Question

I created this regex: /[::].+[^\\>]/g

Test:

 let str = "<foo::bar>" let match = str.match(/[::].+[^\\>]/g).join('') console.log(match)

Expected answer : bar

Actual answer : ::bar

Answers appreciated.

Answer 1

One option is to use a lookbehind assertion ( (?<=) ), which is currently supported only by Chrome & Safari:

 const str = "<foo::bar>" const match = str.match(/(?<=::)[^\\>]+/g).join('') console.log(match)

Answer 2

Adding an extra match & join to remove '::' had solved the problem.

Working code:

let str = "<foo::bar>"
let match = str.match(/[::].+[^\>]/g).join('')
            .match(/[^(::)].+/g).join()
console.log(match)

Answer 3

Another option is to use a capturing group instead of a lookbehind:

::([^>]+)

Regex demo

For example

 const regex = /::([^>]+)/g; const str = `<foo::bar>`; let m; let match = []; while ((m = regex.exec(str)) !== null) { // This is necessary to avoid infinite loops with zero-width matches if (m.index === regex.lastIndex) { regex.lastIndex++; } match.push(m[1]); } console.log(match.join(''));

---

If you want to match the whole pattern of the string, you might use:

<[^>]+::([^>]+)>

Regex demo

 const regex = /<[^>]+::([^>]+)>/g; const str = `<foo::bar>`; let m; let match = []; while ((m = regex.exec(str)) !== null) { // This is necessary to avoid infinite loops with zero-width matches if (m.index === regex.lastIndex) { regex.lastIndex++; } match.push(m[1]); } console.log(match.join(''));