ajax取不到radio的值怎么办

Question

如图，为什么取不到radio的所选值啊

Answer 1

var name = "$("#business_type")".replace("_", "$");
var val = $("input[name='" + name + "']:checked").val();
if (val != undefined) {
switch (val) {
case "1":
break;
case "2":
....
}}