function parseUrl(url){
    let urlParam = url.split("?")[1];
    console.log("---------> URL param : " + urlParam);
    urlParam = urlParam.split("&");
    let urlParamObject = {};
    for(let i=0;i < urlParam.length;i++){
        let tmp = urlParam[i].split("=");
        urlParamObject[tmp[0]] = tmp[1];
    }
    return urlParamObject;
}

let param = parseUrl(url);
param.a // output 10
param.b // output 20

window.location.search.slice(1).split('&').reduce((res, val) => ({...res, [val.split('=')[0]]: val.split('=')[1]}), {})

使用道场。这里没有其他解决方案是如此简短或经过充分测试的：

require(["dojo/io-query"], function(ioQuery){
    var uri = "www.test.com/t.html?a=1&b=3&c=m2-m3-m4-m5 ";
    var query = uri.substring(uri.indexOf("?") + 1, uri.length);
    var queryObject = ioQuery.queryToObject(query);
    console.log(queryObject.c); //prints m2-m3-m4-m5
});

我尽可能喜欢速记：

网址：example.com/mortgage_calc.htm？pmts=120&intr=6.8&prin=10000

香草Javascript：

for ( var vObj = {}, i=0, vArr = window.location.search.substring(1).split('&');
        i < vArr.length; v = vArr[i++].split('='), vObj[v[0]] = v[1] ){}
// vObj = {pmts: "120", intr: "6.8", prin: "10000"}

这是有效的：

function getURLParameter(name) {
  return decodeURIComponent((new RegExp('[?|&]' + name + '=' + '([^&;]+?)(&|#|;|$)').exec(location.href) || [null, ''])[1].replace(/\+/g, '%20')) || null;
}

我没有得到任何其他最好的答案。

您可以运行此函数

    function getUrlVars()
    {
        var vars = [], hash;
        var hashes = window.location.href.slice(window.location.href.indexOf('?') + 1).split('&');
        for(var i = 0; i < hashes.length; i++)
        {
            hash = hashes[i].split('=');
            vars.push(hash[0]);
            vars[hash[0]] = hash[1];
        }
        return vars;
    }

    var source = getUrlVars()["lm_supplier"];
    var el = source.toString();
    var result= decodeURI(el);

console.log(result)

该函数从url获取所需内容，var source=getUrlVars（）[“从url中输入您想要的内容”]；