截取url的参数

gup(name) {
name = name.replace(/[[]/, "\[").replace(/[]]/, "\]");
var regexS = "[\?&]" + name + "=([^&#]*)";
var regex = new RegExp(regexS);
var results = regex.exec(window.location.href);
if (results == null)
return "";
else
return results[1];
},

获取时调用即可