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SpringMVC从Control中响应json数据

2015-12-31 10:39 513 查看
在页面上异步获取Controller中响应的json数据。

<%@ 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>User List</title>

<script type="text/javascript" src="${pageContext.request.contextPath}/resource/script/jquery-1.9.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#button_getUser").click(function(){
//异步请求json数据
$.ajax({
type:"POST",
url:"${pageContext.request.contextPath}/user/getUserList",
success:function(data){
//迭代返回的json数据
$.each(data,function(i,user){
$("#results").append(user.userName+"---"+user.password+"<br>");
});
},
error:function(e) {
alert("出错:"+e);
}
});
});
});
</script>

</head>
<body>
<input type="button" id="button_getUser" value="获取用户列表"><br>
用户信息:<br>
<div id="results">

</div>
</body>
</html>

Controller中的代码:

@RequestMapping("/getUserList")
public String getUserList(HttpServletResponse response) {
response.setCharacterEncoding("UTF-8");
response.setContentType("application/json");

List<User> userList = getUsers();
Gson gson = new Gson();
String json = gson.toJson(userList);
System.out.println("json---"+json);
PrintWriter out = null;
try {
out = response.getWriter();
out.write(json);
out.flush();
}catch(Exception e) {
e.printStackTrace();
}finally {
if(out != null) {
out.close();
}
}

return "/user/userList";
}

private List<User> getUsers() {
List<User> users = new ArrayList<User>();

User user1 = new User();
user1.setUserName("李坏");
user1.setPassword("LiHuai");

User user2 = new User();
user2.setUserName("阿飞");
user2.setPassword("aFei");

User user3 = new User();
user3.setUserName("陆小凤");
user3.setPassword("LuXiaoFeng");

users.add(user1);
users.add(user2);
users.add(user3);

return users;
}


或者使用@ResponseBody,需要用到jackson相关的包。
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