URL 正则表达式模式

Question

I would like to parse URLs with Regular Expressions and find a pattern that matches with https://*.global .

Here is my URL test string on regex101.

Ideally, the regex would return https://app8.global instead of cover other https string.

 const URL = `https://temp/"https://app8.global"https://utility.localhost/`; const regex = /https:\\/\\/(.+?)\\.global(\\/|'|"|`)/gm; const found = URL.match(regex); console.log(found);

How would I manipulate the regex so it will return the https://*.global ?

Answer 1

First of all, you need to exclude slashes from the starting part, otherwise it'll match things from the previous url:

const regex = /https:\/\/([^\/]+?)\.global(\/|'|"|`)/gm;

Now, you can convert the weird 4 character or with a character group:

const regex = /https:\/\/([^\/]+?)\.global[\/'"`]/gm;

And now you can get the matches and trim off that last character:

const matches = URL.match(regex).map(v => v.slice(0, -1));

Then, matches would evaluate to ["https://app8.global"] .

Answer 2

Using Group RegExp.$1

 const URL = `https://temp/"https://app8.global"https://utility.localhost/`; const regex = /(https:\\/\\/([^\\/]+?)\\.global[\\/'"`])/; const found = URL.match(regex); console.log(RegExp.$1);