jqGrid Server端读取数据展示(Java Servlet)

package com.polaris.jqgrid.servlet;

import java.io.IOException;

import javax.servlet.ServletException;
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;
/**
* 该Servlet向客户端返回一个json对象。为了简便，数据不是从数据库获得的。
* jqGrid默认期望返回的json对象格式要求如下：
* {"page":"1","total":"2","records":"13",
* "rows":[
*???????????????? {id:"1",cell:["1","polaris","男","polaris@gmail.com","772618379","18329382732","1985-10-2"]},
*???????????????? {id:"2",cell:["2","张三","女","zhangsan@163.com","272618382","15329382732","1986-10-12"]},
*???????????????? {id:"3",cell:["3","王五","女","wangwu@yahoo.com","172635372","13329389832","1987-12-21"]},
*???????????????? {id:"4",cell:["4","赵六","男","zhaoliu@sina.com","372618332","18929343731","1988-09-22"]}
*???????? ]
* }
* 当然，在js中，可以通过jqGrid的jsonReader属性来修改默认格式
* 因为默认的格式，rows的数据要求顺序不能变，且每个字段都得有值（空也得有"")。因而，
* 在jsonReader中定义repeatitems : false。这样，rows就变成了：
* "rows":[
*???????????????? {id:"1",userName:"polaris",gender:" 男",email:"polaris@gmail.com",QQ:"772618379",mobilePhone:"18329382732",birthday:"1985-10-2"]},
*???????????????? {id:"2",userName:"徐新华",gender:" 男",email:"xh.xu@163.com",QQ:"272618382",mobilePhone:"15329382732",birthday:"1986-10-12"]},
*???????????????? {id:"3",userName:"王五",gender:" 女",email:"wangwu@yahoo.com",QQ:"172635372",mobilePhone:"13329389832",birthday:"1987-12-21"]},
*???????????????? {id:"4",userName:"赵六",gender:" 女",email:"zhaoliu@sina.com",QQ:"372618332",mobilePhone:"18929343731",birthday:"1988-09-22"]}
*???????? ]
* @author xuxinhua
*
*/
public class JqGridForJSONServlet extends HttpServlet
{
????????private static final long serialVersionUID = 132383828833L;

????????@Override
????????public void doGet(HttpServletRequest req, HttpServletResponse resp)
????????????????????????throws ServletException, IOException {
????????????????// 定义返回的数据类型：json，使用了json-lib
????????????????JSONObject jsonObj = new JSONObject();
????????????????// 根据jqGrid对JSON的数据格式要求给jsonObj赋值
????????????????jsonObj.put("page", 1);????????????????// 当前页
????????????????jsonObj.put("total", 1);????????// 总页数
????????????????jsonObj.put("records", 4);????????// 总记录数
????????????????// 定义rows，存放数据
????????????????JSONArray rows = new JSONArray();
????????????????// 放入4条数据
????????????????for(int i=0;i<4;i++)
????????????????{
????????????????????????// 存放一条记录的对象
????????????????????????JSONObject cell = new JSONObject();
????????????????????????cell.put("id", i);
????????????????????????if(i%2==0)
????????????????????????{
????????????????????????????????cell.put("userName", "polaris");
????????????????????????????????cell.put("gender", "女");
????????????????????????}
????????????????????????else
????????????????????????{
????????????????????????????????cell.put("userName", "徐新华");
????????????????????????????????cell.put("gender", "男");
????????????????????????}
????????????????????????cell.put("email", "polaris@gmail.com");
????????????????????????cell.put("QQ", "772"+i+"1837"+i);
????????????????????????cell.put("mobilePhone", "132"+i+"1837"+i+"3"+i);
????????????????????????cell.put("birthday", "198"+i+"-10-"+"1"+i);
????????????????????????// 将该记录放入rows中
????????????????????????rows.add(cell);
????????????????}
????????????????// 将rows放入json对象中
????????????????jsonObj.put("rows", rows);
????????????????// 自控制台打印输出，以检验json对象生成是否正确
????????????????System.out.println("要返回的json对象：\n" + jsonObj.toString());
????????????????// 设置字符编码
????????????????resp.setCharacterEncoding("UTF-8");
????????????????// 返回json对象（通过PrintWriter输出）
????????????????resp.getWriter().print(jsonObj);
????????}
????????@Override
????????public void doPost(HttpServletRequest req, HttpServletResponse resp)
????????????????????????throws ServletException, IOException {
????????????????doGet(req,resp);
????????}
}