Jquery form 提交success返回值ALERT出现404HTML代码

Question

Jquery 获取后台返回值异常。如下图：不知道为什么会这样，在另一个页面也写了一个一样的返回接收都是正常的。

function saveUser(){
			var lnum = $("#lnumber").val();
			var needc = $("#needcount").val();
			var needd = $("#hddate").val();
			
			var errorMsg = "";
			if(lnum == ""||lnum == null){
				errorMsg += "请输入料号！rn";
			}
			if(needc == ""||needc == null){
				errorMsg += "请输入需求数量！rn";
			}
			if(needd == ""||needd == null){
				errorMsg += "请选择需求日期！rn";
			}
			if(errorMsg == ""){
				document.getElementById("hidtype").value="submit";
				$("#uploadform").form("submit", {
                	url: "demand/addDemand",
                	type:"GET",
                	dataType:"JSON",
                	onsubmit: function () {
                    	return $(this).form("validate");
                	},
                	success: function (result) {
                	alert(result);
                    	if (result == "1") {
                        	$.messager.alert("提示信息", "提交成功");
                    	}
                    	else {
                        	$.messager.alert("提示信息", "提交失败");
                    	}
                	}
            	});
			}else{
				$.messager.alert("提示信息",errorMsg);
			}
		}

String lnumber = request.getParameter("lnumber");
		String needcount = request.getParameter("needcount");
		String needdate = request.getParameter("needdate");
		String needdescribe = request.getParameter("needdescribe");
		String needurl = JSON.toJSONString(list);
		
		int num;
		Demand demand = new Demand();
		Map<String, Object> map = new HashMap<String, Object>();
		
		SimpleDateFormat sdf = new SimpleDateFormat("yyyy-mm-dd");
		try {
			demand.setLnumber(lnumber);
			demand.setNeedcount(Integer.parseInt(needcount));
			demand.setNeeddate(sdf.parse(needdate));
			demand.setNeeddescribe(needdescribe);
			demand.setNeedurl(needurl);
			num = this.demandService.addDemand(demand);
			map.put("num", num);
		} catch (Exception e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
			map.put("num", "error");
			throw e;
		} 
//		return JSONArray.toJSONString(map).toString();
		return String.valueOf(num);

Answer 1

怎么解决的

Answer 2

已解决