求助：提取url参数

Question

从这个链接里面提取后面的两个参数，然后再保存到本地let href = 'http://w3c.pcidata.cn/pci_int...'

Answer 1

function queryParse (str) {
  return str.split('&').reduce((res, kv) => {
    const [k, v] = kv.split('=');
    res[k] = decodeURIComponent(v);
    return res;
  }, {});
}

const QUERY = queryParse(location.search.slice(1));

简易的实现，同名参数是直接覆盖的（一般够用）

Answer 2

let hashString = (new URL('http://w3c.pcidata.cn/pci_integral_mall_h5/#/?goodsType=09&mediaId=mt2000000019')).hash;
// #/?goodsType=09&mediaId=mt2000000019
let params = new URLSearchParams(hashString.replace('#/?',''));// hash 去掉 #/?
// [["goodsType", "09"],["mediaId", "mt2000000019"]]
console.log(params.get('mediaId'));// 获取参数
// mt2000000019
console.log(Object.fromEntries(params));// 数组转对象
// {goodsType: "09", mediaId: "mt2000000019"}

注意：不兼容IE

https://developer.mozilla.org...
https://developer.mozilla.org...
https://developer.mozilla.org...

Answer 3

function getQueryVal(url){
    url = url || '';
    let arr = (url.split("?")[1] || url || '').split("&");
    let o={};
    if(arr[0]==''){return o}
    arr.forEach(function(x){
        let ar=x.split("=");
        o[ar[0]]=ar[1]
    });
    return o
}