我正在發出一個AJAX請求並將param發送給servlet。我可以看到參數的URL，但我沒能獲得請求參數在servlet

Question

這是我的jQuery代碼

//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> 

的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(); 
     } 

    } 

Answer 1

你必須在你的servlet上有這樣的東西：

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

其中我不能在你的代碼中看到。