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)

 

posted @ 2012-05-30 14:23  WebApi  阅读(352)  评论(0编辑  收藏  举报
CopyRight © 博客园 WebAPI