怎么过滤url上的xss代码

[wulucxy]

类似于这种路由地址：

https://www.host.com/app/follow?follow=1&uid=2300052</script+><script>alert(583)</script>

[Lizhooh]

var xss = require('xss');
var url = xss('https://www.host.com/app/follow?follow=1&uid=2300052</script+><script>alert(583)</script>');
console.log(url);