Question

我是新手，所以我没有多少经验。我的问题很简单，我想在我的表中用ajax加载数据 - 但我不知道我该怎么做。我现在有这个代码（ajaxServlet）：

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    int row=0;
    JSONArray back = new JSONArray();
    JSONObject obj = new JSONObject();
    try{
        Class.forName("com.jdbc.mysql.Driver");
        Connection con = DriverManager.getConnection("jdbc:mysql//localhost/expenses","***","****");
        PreparedStatement stmt = con.prepareStatement("Select * from expenses");
        ResultSet rs = stmt.executeQuery();


        while(rs.next())
            {
                JSONArray array = new JSONArray();
                array.add(rs.getString("UID"));
                array.add(rs.getString("Last Name");
                array.add(rs.getString("First Name"));

                back.add(array);
                row++;
            }

        PrintWriter out = response.getWriter();
        response.setContentType("application/json");
        obj.put("size", row); // Here  is located the problem i think...what must i return from here?
        obj.put("example", back);
        out.println(obj);
    }catch(Exception e)
    {
        e.printStackTrace();
    }
}

在HTML方面，代码与documentaion

中的代码完全相同

只有ajaxUrl看起来像这样：

ajaxUrl : "ajaxServlet?page={page}&size={size}&{sortList:col}",

所以，请你告诉我，我必须改变或做什么，这是有效的。

非常感谢

Ajax加载不起作用

0 个答案: