javascript - 如何使用 AJAX 从 jsp 中的 servlet 检索多个值

Question

我有这个 ajax javascript 代码调用 servlet 来检索两个值(名字、电话)。我知道如何从 servlet 获取单个值而不是多个值。

这是我的ajax

    <script>
        function getCustomerDetailsAjax(str) {
            str = $('#customerId').val();

            if (document.getElementById('customerId').value <= 0) {
                document.getElementById('firstName').value = " ";
                document.getElementById('telephone').value = " ";
                document.getElementById('vehicleMake').value = " ";
                document.getElementById('vehicleModel').value = " ";
                document.getElementById('vehicleColor').value = " ";
            } else {
                $.ajax({
                    url: "GetCustomerDetails",
                    type: 'POST',
                    data: {customerId: str},
                    success: function (data) {                       
                        alert(data); //I want to get 2 servlet values and alert them here. How can I do that?
                    }
                });
            }
        }
    </script>

这是我的servlet

public class GetCustomerDetails extends HttpServlet {

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    PrintWriter out=response.getWriter();
    int customerId = Integer.valueOf(request.getParameter("customerId"));
    try {
        Class.forName("com.mysql.jdbc.Driver");
        Connection con = DriverManager.getConnection("jdbc:mysql://localhost/Vehicle", "root", "");
        PreparedStatement ps = con.prepareStatement("SELECT fistname,telephone FROM customers WHERE customerid=?");
        ps.setInt(1, customerId);
        ResultSet result=ps.executeQuery();
        if(result.next()){
            out.print(result.getString("firstname")); //I want to send this value
            out.print(result.getString("telephone")); //and this value

        }

    } catch (ClassNotFoundException ex) {
        Logger.getLogger(GetCustomerDetails.class.getName()).log(Level.SEVERE, null, ex);
    } catch (SQLException ex) {
        Logger.getLogger(GetCustomerDetails.class.getName()).log(Level.SEVERE, null, ex);
    }

}

@Override
public String getServletInfo() {
    return "Short description";
}// </editor-fold>

这是从servlet中获取数据的部分，如何从中获取多个值并发出警报？

       success: function (data) {                       
            alert(data); //I want to get 2 servlet values and alert them here. How can I do that?
       }

谢谢!

Answer 1

要在您的 Web 服务和客户端之间共享数据，您必须选择最适合您需求的协议(protocol)/策略(XML、JSON...)。

由于您使用的是 JavaScript，我建议您阅读有关 JSON(代表“JavaScript 对象表示法”)的内容。

在您的示例中，您应该生成并返回一个 JSON 字符串(带有正确的 Content-type header )- 您可以阅读有关 javax.json 包的信息。使用 JSON，您可以返回包含您选择的字段的数据结构。

类似的东西(未经测试——我已经很久没有编写 Java 代码了):

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    PrintWriter out=response.getWriter();
    int customerId = Integer.valueOf(request.getParameter("customerId"));
    try {
        Class.forName("com.mysql.jdbc.Driver");
        Connection con = DriverManager.getConnection("jdbc:mysql://localhost/Vehicle", "root", "");
        PreparedStatement ps = con.prepareStatement("SELECT fistname,telephone FROM customers WHERE customerid=?");
        ps.setInt(1, customerId);
        ResultSet result=ps.executeQuery();
        if(result.next()){

            /* set response content type header: jQuery parses automatically response into a javascript object */
            response.setContentType("application/json");
            response.setCharacterEncoding("utf-8");

            /* construct your json */
            JsonObject jsonResponse = new JsonObject();
            jsonResponse.put("firstname", result.getString("firstname"));
            jsonResponse.put("telephone", result.getString("telephone"));            

            /* send to the client the JSON string */
            response.getWriter().write(jsonResponse.toString());
           // "{"firstname":"first name from db","telephone":"telephone from db"}"

        }

    } catch (ClassNotFoundException ex) {
        Logger.getLogger(GetCustomerDetails.class.getName()).log(Level.SEVERE, null, ex);
    } catch (SQLException ex) {
        Logger.getLogger(GetCustomerDetails.class.getName()).log(Level.SEVERE, null, ex);
    }

}

在你的 JS 中(我想你正在使用 jQuery，因为 success 回调):

   success: function (data) { 
        /* because you set the content-type header as 'application/json', you'll receive an already parsed javascript object - don't need to use JSON.parse. */


        console.log(data);
        /*
            {
                firstname: "first name from db",
                telephone: "telephone from db"
            }

        */

        alert(data.firstname); //alert firstname
        alert(data.telephone); //alert phone
   }