dojo的ajax使用入门
dojo 怎么样实现和server端的交互？那么它有自己的一套AJAX框架的内容
<%@ page language="java" contentType="text/html; charset=utf-8"
pageEncoding="utf-8"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>dojo ajax</title>
<script type="text/javascript" src="../dojo/dojo/dojo.js" djConfig="parseOnLoad: true"></script>
<script type="text/javascript">
function init(){
var timestamp = Date.parse(new Date());
var json_content={name:"huangbiao",age:14,time:timestamp};
dojo.xhrGet({
url: "../DojoAjax",//发送到servlet的路径
content:json_content,//这个参数是用来传递参数的
load: function(response, ioArgs){//成功响应的事件
//console.log("xhr get success:", response);
alert("xhr get success:", response);
return response;
},
sync:true,//默认是false
error: function(response, ioArgs){//失败响应的
//console.log("xhr get failed:", response);
alert("xhr get failed:", response);
return response;
}
});
}
//dojo.addOnLoad(init);
</script>
</head>
<body>
<form action="../DojoAjax">
<input type="submit" value="submit">
</form>
<button onclick="init();">ajax</button>
</body>
</html>server端的代码
package hb.servlet;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class DojoAjax extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
System.out.println("do get");
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
System.out.println("do post");
}
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