dojox.grid.DataGrid显示数据

?第一种方式直接得到已经存在的数据（存在本地或者是已经写死的JSON对象中），不需要跟server交互

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第二种：数据来源于server端，和server进行交互?
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server端的代码
package hb.servlet;import java.io.IOException;import java.io.PrintWriter;import javax.servlet.ServletException;import javax.servlet.http.HttpServlet;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;/** * Servlet implementation class WriteJson */public class WriteJson extends HttpServlet {private static final long serialVersionUID = 1L;       protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {this.doPost(request, response);}protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {System.out.println("{items:[{name:'Africa', type:'continent'},{name:'Africa1', type:'continent1'}]}");PrintWriter pw = response.getWriter();pw.write("{items:[{name:'Africa', type:'continent'},{name:'Africa1', type:'continent1'},{name:'Africa1', type:'continent1'},{name:'Africa1', type:'continent1'},{name:'Africa1', type:'continent1'}]}");pw.flush();pw.close();}}
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备注：server端输出的字符串必须也要按照json对象中包含items属性的这中格式，否则是无法在前端显示内容的。
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