用Servlet返回JSON文本动态创建DataGrid
Posted 码上加油站
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了用Servlet返回JSON文本动态创建DataGrid相关的知识，希望对你有一定的参考价值。
<%@ page language="java" pageEncoding="UTF-8"%> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <title>用Servlet返回JSON文本动态创建DataGrid</title> <!-- 引入css文件，无顺序 --> <link rel="stylesheet" href="themes/icon.css" type="text/css"></link> <link rel="stylesheet" href="themes/default/easyui.css" type="text/css"></link> <!-- 引入js文件，有顺序 --> <script type="text/javascript" src="js/jquery.min.js"></script> <script type="text/javascript" src="js/jquery.easyui.min.js"></script> <script type="text/javascript" src="js/easyui-lang-zh_CN.js"></script> </head> <body> <table id="dg"></table> <script type="text/javascript"> $("#dg").datagrid({ url : "${pageContext.request.contextPath}/UserServlet?time="+new Date().getTime(), columns : [[ {field:‘id‘,title:‘编号‘,width:100}, {field:‘name‘,title:‘姓名‘,width:100}, {field:‘sal‘,title:‘薪水‘,width:100}, {field:‘sex‘,title:‘性别‘,width:100} ]], fitColumns : true, singleSelect : true, pagination : true, pageSize : 2, pageList : [2,4] }); </script> </body> </html>
package loaderman.bean2json; import java.io.IOException; import java.io.PrintWriter; import java.util.ArrayList; import java.util.LinkedHashMap; import java.util.List; import java.util.Map; import javax.servlet.ServletException; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import net.sf.json.JSONArray; /** * 用Servlet返回JSON文本动态创建DataGrid * * @author AdminTC */ public class UserServlet extends HttpServlet { public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { this.doPost(request, response); } public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { request.setCharacterEncoding("UTF-8"); //收集DataGrid向服务器发送的参数--page(当前页号) String page = request.getParameter("page"); System.out.println("page=" + page); //收集DataGrid向服务器发送的参数--rows(当前需要显示的记录数) String rows = request.getParameter("rows"); System.out.println("rows=" + rows); List<User> userList = new ArrayList<User>(); userList.add(new User(1, "哈哈", 1000, "男")); userList.add(new User(2, "呵呵", 2000, "男")); userList.add(new User(3, "嘻嘻", 3000, "女")); userList.add(new User(4, "笨笨", 4000, "男")); userList.add(new User(5, "聪聪", 5000, "男")); userList.add(new User(6, "月月", 6000, "女")); userList.add(new User(7, "花花", 7000, "女")); Map<String, Object> map = new LinkedHashMap<String, Object>(); map.put("total", userList.size()); map.put("rows", userList); JSONArray jsonArray = JSONArray.fromObject(map); String jsonJAVA = jsonArray.toString(); jsonJAVA = jsonJAVA.substring(1, jsonJAVA.length() - 1); //以流的方式将JSON文本输出到DateGrid组件中 response.setContentType("text/html;charset=UTF-8"); PrintWriter pw = response.getWriter(); pw.write(jsonJAVA); pw.flush(); pw.close(); } }
以上是关于用Servlet返回JSON文本动态创建DataGrid的主要内容，如果未能解决你的问题，请参考以下文章
用jquery的post方法提交数据需要servlet返回的数据类型是json的数据格式，应该怎么做？？