Java 如何从AJAX数据获取响应消息

在这里，我编写对ajax响应对象的响应

protected void writeAjaxResponse(HttpServletRequest req 
                                 ,HttpServletResponse resp,String result){
        PrintWriter writer = null;
        try {
            writer = resp.getWriter();
        } catch (IOException e) {
            e.printStackTrace();
        }
        writer.println(result);
        return;

    } 

后来我打电话来

writeAjaxResponse(req, resp, "<p style=color:red>Error occured recording
                                                           your feedback!</p>");

但我正在戒备

    [object XMLDocument]

编辑：

这是我的servlet

doPost（）

方法

@Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException { 

         String feedbacker = req.getParameter("feedbacker");
         String  feedbackeremail = req.getParameter("feedbackeremail");
         String  feedbackermsg = req.getParameter("feedbackermsg");

        boolean saveFeedback = MailSenderServlet.
            saveFeedback(req, resp, feedbackeremail, "",
                  feedbackermsg, feedbacker, feedbackeremail);
        if(saveFeedback){
            writeAjaxResponse(req, resp, "Feedback received succesfully!");
        }else{
            writeAjaxResponse(req, resp, "Error occured  !");
        }

    } 

但我期待着我的回复

如果我错过了什么，请告诉我

请帮忙

再注射30分钟后

我发现缺少

MIME类型

，并将方法更改为

protected void writeAjaxResponse(HttpServletRequest req 
                                 ,HttpServletResponse resp,String result){
       resp.setContentType("text/html;charset=UTF-8");
        PrintWriter writer = null;
        try {
            writer = resp.getWriter();
        } catch (IOException e) {
            e.printStackTrace();
        }
        writer.println(result);
        return;

    }

---

感谢@Noob@w4rumy@user2207792提供的提示支持。

如果您使用的是Chrome，您可以使用

console.log（data）

并检查控制台以了解您的响应对象的确切外观。数据类型：'html'//或json或其他什么？在ajax中！我在firebug和chrome中看到了空控制台：（

dataType:JSON，

in$.ajax！！！@NoobUnChained是的，谢谢。让我检查一下。

protected void writeAjaxResponse(HttpServletRequest req 
                                 ,HttpServletResponse resp,String result){
       resp.setContentType("text/html;charset=UTF-8");
        PrintWriter writer = null;
        try {
            writer = resp.getWriter();
        } catch (IOException e) {
            e.printStackTrace();
        }
        writer.println(result);
        return;

    }