获取地址栏URL中参数, getQuerySting()方法

今天同事用的以前的获取url地址参数获取不到.以前的方法失效了.后面发现是正则表达式bug:

第一种获取方法(针对普通情况的一般够用):

function getQueryString(name) {
    var reg = new RegExp("(^|&)" + name + "=([^&]*)(&|$)", "i");
    var reg_rewrite = new RegExp("(^|/)" + name + "/([^/]*)(/|$)", "i");
    var r = window.location.search.substr(1).match(reg);
    var q = window.location.pathname.substr(1).match(reg_rewrite);
    if(r != null){
        return unescape(r[2]);
    }else if(q != null){
        return unescape(q[2]);
    }else{
        return null;
    }
}
第二种获取方法:
这样的url地址获取不到:file:///Users/wanghong/work/ziku/youle/dist/index.html?sourceType=Tsingtao_beer_activity#/welcome?toPath=file%3A%2F%2F%2FUsers%2Fwanghong%2Fwork%2Fziku%2Fyoule%2Fdist%2Findex.html%3FsourceType%3DTsingtao_beer_activity%23%2F

/**
 * 获取指定的URL参数值
 * URL:http://www.quwan.com/index?name=tyler
 * 参数：paramName URL参数
 * 调用方法:getParam("name")
 * 返回值:tyler
 */
function getParam(paramName) {
    paramValue = "", isFound = !1;
    if (this.location.search.indexOf("?") == 0 && this.location.search.indexOf("=") > 1) {
        arrSource = unescape(this.location.search).substring(1, this.location.search.length).split("&"), i = 0;
        while (i < arrSource.length && !isFound) arrSource[i].indexOf("=") > 0 && arrSource[i].split("=")[0].toLowerCase() == paramName.toLowerCase() && (paramValue = arrSource[i].split("=")[1], isFound = !0), i++
    }
    return paramValue == "" && (paramValue = null), paramValue
}

大家都尝试一下.元宵节快乐

3月5号后端了发现bug
我们的URL地址:http://localhost:8081/#/welcome?toPath=http%3A%2F%2Flocalhost%3A8081%2F%23%2F%3FsourceType%3DTsingtao_beer_activity
还是获取不到sourceType参数
咨询了群里的守候,,提供了getUrlPromt()了方法

function getUrlPrmt(url) {

url = url ? url : window.location.href;

let _pa = url.substring(url.indexOf('?') + 1), _arrS = _pa.split('&'), _rs = {};

for (let i = 0, _len = _arrS.length; i < _len; i++) {

let pos = _arrS[i].indexOf('=');

if (pos == -1) {

continue;

}

let name = _arrS[i].substring(0, pos), value = window.decodeURIComponent(_arrS[i].substring(pos + 1));

_rs[name] = value;

}

return _rs;

}

name我这里的方法写法应该写成这样getUrlPrmt(getUrlPrmt(location.href).toPath).sourceType

vue中的跳转写法:

this.$router.push({ path: '/register', query: { sourceType: window.getUrlPrmt(window.getUrlPrmt(location.href).toPath).sourceType }

谢谢守候大牛的方法.