js删除url参数并返回url

javascript 删除 url 中指定参数,并返回 url

// 删除url中某个参数,并跳转

function funcUrlDel(name){    var loca = window.location;    var baseUrl = loca.origin + loca.pathname + "?";    var query = loca.search.substr(1);    if (query.indexOf(name)>-1) {        var obj = {}        var arr = query.split("&");        for (var i = 0; i < arr.length; i++) {            arr[i] = arr[i].split("=");            obj[arr[i][0]] = arr[i][1];        };        delete obj[name];        var url = baseUrl + JSON.stringify(obj).replace(/[\"\{\}]/g,"").replace(/\:/g,"=").replace(/\,/g,"&");        return url    };}

功能:删除url中指定的参数,并返回删除参数后的完整url

使用方法

示例

url: http//xx.com/list?page=1&a=5

执行代码

funcUrlDel("page")

返回

http//xx.com/list?a=5