在jquery中返回错误响应

时间:2016-06-27 06:15:13

标签: java jquery ajax jsp

我正试图从jquery调用一个控制器。这里的每件事都运行正常,但是在ajax中它会显示错误提示。出了什么问题。

$.ajax({
                         type: 'POST',
                         url:'<%=request.getContextPath()%>/billing',
                         data: ({caseId: caseId,  noteID :  noteID}),    
                         cache: false,
                         success: function(data){
                                alert("Success");
                            },
                         error: function(xhr, textStatus, error){
                         alert("Error occured.");
                         }
                         });

我的控制器

@RequestMapping(value = "/billing", method = RequestMethod.POST)
    public String Billing(@RequestParam Long caseId,
            @RequestParam Long noteID, HttpServletRequest request)  throws Exception {
        try{

           ----Some data base updation---
            logger.debug("success ");
        return "success";
        } catch (Exception e) {
            logger.error(e,e);
            throw e;
        }}

请帮助我。

1 个答案:

答案 0 :(得分:0)

//你必须离开内容类型才能发送服务器端,请定义内容类型..

 $.ajax({
                type: "POST",
                url: '@Url.Action("PerTGradeMastEdit", "Master")',
                data: "{'Request':'" + request + "'}",
                contentType: "application/json; charset=utf-8",
                success: function (response) {
                    if (response != null) {

                        var strResponse = response;
                        var ProgramData = strResponse.split("/");
                        // response = ProgramData[0];programId
                        $('#txtPerGrade').val(ProgramData[0]);
                        $('#txtDesc').val(ProgramData[1]);

                    }
                },
                failure: function (msg) {
                    alert(msg);
                    window.location.href = '@Url.Action("INDEX", "AUTHORIZE")';
                }
            });`enter code here`