js获取url参数(通用方法)

function getUrl(name="") {
    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]]=unescape(strs[i].split("=")[1]);
        }
    }
   if(name == ""){
      return theRequest;
   }else{
      return theRequest[name]?theRequest[name]:'';
   }
}

用法:getUrl("id");

注意

  1.不传入值则返回地址的所有参数

  2.若没有找到匹配的参数则返回一个空的JSON

原文地址:https://www.cnblogs.com/zhizou/p/10488125.html