[英]How to get response message from AJAX data
在這里我將響應寫到ajax響應對象
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;
}
后來我打電話
writeAjaxResponse(req, resp, "<p style=color:red>Error occured recording
your feedback!</p>");
在jQuery中
$.ajax({
type: 'POST',
url: 'savefeedback',
data: 'feedbacker='+feedbacker+'feedbackeremail=
'+feedbackeremail+'feedbacker='+feedbackermsg,
success:function(data){
alert(data); //here is the pin point
}
});
但是在戒備狀態下
[object XMLDocument]
編輯:
這是我的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 !");
}
}
但是我期待着我的回復消息。
如果我錯過了什么,請告訴我。
請幫忙!!!!
30分鍾后重新研磨
我發現缺少MIME type
,並將方法更改為
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;
}
感謝@Noob @ w4rumy @ user2207792的迅速支持。
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