js获取url参数
js获取url参数,参数在?后边
方法1:
function QueryString(item) { var sValue=location.search.match(new RegExp("[\?\&]"+item+"=([^\&]*)(\&?)","i")) return sValue?sValue[1]:sValue } //使用方法 //url=http://www.xxx.com?ID=100&name=dom var ID=QueryString("ID"); var name=QueryString("name");
方法2:
function GetRequest() { var url = location.search; //获取url中"?"符后的字串 var theRequest = new Object(); if (url.indexOf("?") != -1) { var str = url.substr(1); strs = str.split("&"); for (var i = 0; i < strs.length; i++) { theRequest[strs[i].split("=")[0]] = decodeURIComponent(strs[i].split("=")[1]); } } return theRequest; } var RequestRef = GetRequest()["r"]; console.log(RequestRef)
//成功一定有方法,失败一定有原因。