添加参数后,浏览器找不到servlet网址,500找不到问题页面?result =

时间:2018-04-10 14:54:42

标签: javascript java jsp

在我使用没有参数的$.get("QuestionPage"...之前,一切都很好。我改成了:

$.get("QuestionPage",{result:"nailed"},function(responseJSON){

           //set the question and category texts
           $("#cardArea").val(responseJSON["card"]);
           $( "#cat"  ).text(responseJSON["category"]);
           $( "#cardid" ).text(responseJSON["flashCardnum"]);

           //clear answer field after every 'next'...
           $( "#answerArea" ).val("");

           //set the click function of show button to show the answer..
           $( "#show" ).click(function() {
                $( "#answerArea" ).val(responseJSON["answer"]); 
                answerShown = true;
            });
       });

现在我得到500错误无法加载资源。这是servlet方法的一部分:

@WebServlet("/QuestionPage")
public class QuestionPageServlet extends HttpServlet {

    private Map<String,UserAccount> activeUsers;
    private List<Card> usersUpdatedCards = new ArrayList<>();  

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {

1 个答案:

答案 0 :(得分:0)

您忘记了ajax请求中的$.param变量。

$.get("../QuestionPage", $.param({result:"nailed"}), function(responseJSON) {