Question

我有以下代码：

package server;

import java.io.IOException;
import java.io.PrintWriter;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.ResultSetMetaData;

import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import net.sf.json.JSONArray;
import net.sf.json.JSONObject;

import org.json.simple.JSONValue;


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


public servlet() {
    super();

}


protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
doGet(request, response);
}


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

     response.setContentType("application/json");
     PrintWriter out = response.getWriter();

     JSONObject responseData=new JSONObject();       
         JSONArray jArray=new JSONArray();


    try {

        Class.forName("com.mysql.jdbc.Driver");

        Connection con = DriverManager.getConnection("jdbc:mysql://localhost:8080/neue","*****","****");
        PreparedStatement prepared = con.prepareStatement("Select * from person");
        ResultSet rs = prepared.executeQuery();

        ResultSetMetaData rsmd = rs.getMetaData();
        int numberOfColumns = rsmd.getColumnCount();



        while(rs.next())
        {
            JSONObject jObject=new JSONObject();
            JSONArray Jarr = new JSONArray();
            Jarr.add(rs.getString("a"));
            Jarr.add(rs.getString("b"));
            Jarr.add(rs.getString("c"));
            Jarr.add(rs.getString("d"));
            Jarr.add(rs.getString("e"));
                jObject.put("cell", Jarr);
                jArray.add(jObject);
        }
         responseData.put("total", 5);
             responseData.put("page", 1);
             responseData.put("records",numberOfColumns);

         responseData.put("rows",jArray);

         out.println(JSONValue.toJSONString(responseData));

    } catch (Exception e) {


    }
}

}

Script.js看起来像这样：

$(document).ready(function (){
$("#list").jqGrid({
    url:'servlet',
    datatype: "json",
    mtype : 'get',
    colNames:['a','b', 'c', 'd','e'],
    colModel:[
        {name:'a',index:'a', width:75},
        {name:'b',index:'b', width:90},
        {name:'c',index:'c', width:100},
        {name:'d',index:'d', width:80,align:"right"},
        {name:'e',index:'e', width:80, align:"right"},
    ],

    rowNum:10,
    autowidth: true,
    rowList:[10,20,30],
    pager: jQuery('#pager'),
    sortname: 'id',
    viewrecords: true,
    sortorder: "desc",
    caption:"JSON Example",
    jsonReader: {
        repeatitems: false,
        id: "0",
        cell: "",
        root: "rows",
        page: function() { return 1; },
        total: function() { return 1; },
        records: function(obj) { return obj.length; }
    }
}).navGrid('#pager',{edit:false,add:false,del:false});

}）;

正如您所见，我尝试从servlet获取数据。但我没有得到他们的回应。

表格仍然是空的，只是我能看到的标题。在firebug控制台中，我收到了这条消息：

状态代码：500内部服务器错误

我该怎么做才能运行该程序？

非常感谢

JQGRID Servlet-MySQL无法正常工作

0 个答案: