I'm making an AJAX request and sending param to servlet. I can see parameter in the url but i'm not able to get the request parameter in the servlet

Question

This is my jquery code

//jquery
<script>
$.post("dashboardcon", $("#date").serialize(), function(responseHtml) {
    $('#date').html(responseHtml); 
});
</script>

html

<form>
    Date :<br> <input id="date" type="text" name="date">
         <input type="submit" value="Submit">
</form>

servlet

@WebServlet("/dashboardcon")
public class dashboardCon extends HttpServlet {
    private static final long serialVersionUID = 1L;

    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        JSONObject jObj = new JSONObject();
        try {
            JSONObject newObj = jObj.getJSONObject(request.getParameter("date"));
            System.out.println(newObj);
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }

Are you sure you're sending to the correct URL of your servlet? — Difster, Jul 27 '17 at 08:09
What is the back end language and how are you checking for the parameter there? — Difster, Jul 27 '17 at 08:13
You are using ajax request to sending form data of jquery Post method. You should use $.post( "dashboardcon", { date: $("#date").val()},funcion.. — Nirdesh Sharma, Jul 28 '17 at 10:33

score 0 · Answer 1 · answered Jul 27 '17 at 08:14

0

You must have something like this on your servlet:

 JSONObject jObj    = new JSONObject();
 JSONObject newObj = jObj.getJSONObject(request.getParameter("date"));

Whichi i cant see in your code.

answered Jul 27 '17 at 08:14

Fotis Grigorakis

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now I'm is getting this error "org.json.JSONException: Null key. null...." – sandeep balaji Jul 27 '17 at 08:22
if you don't post your code i can't do anything by guessinh that! You must help us for helping you! Thats the way it goes. – Fotis Grigorakis Jul 27 '17 at 08:24
1

I'm sorry! I added servlet code also – sandeep balaji Jul 27 '17 at 08:32

I'm making an AJAX request and sending param to servlet. I can see parameter in the url but i'm not able to get the request parameter in the servlet

1 Answers1