我有一个配置文件,如何找到所有的js URL?js URL像这样:https://raw.githubusercontent.com/xx/Script/master/File/xxx.jshttps://raw.githubusercontent.com/xx/Script/master/File/rrrxxxrrr.js
并将URL替换为js文件,如下所示:
https://raw.githubusercontent.com/xx/Script/master/File/xxx.js至xxx.jshttps://raw.githubusercontent.com/xx/Script/master/File/rrrxxxrrr.js到rrrxxxrrr.js
谢谢。
如果您感兴趣的是替换所有内容,直到URL的最后一个/
,您可以使用以下表达式:/https?:\/\/[^\s]+\//gi
(用空字符串替换)。
示例:
const exp = /https?:\/\/[^\s]+\//gi;
const input = `https://raw.githubusercontent.com/xx/Script/master/File/xxx.js
https://raw.githubusercontent.com/xx/Script/master/File/rrrxxxrrr.js
https://raw.githubusercontent.com/xx/Script/master/File
http://raw.githubusercontent.com/rrrxxxrrr.js`;
console.log(input.replace(exp, ''));