Home > Article > Web Front-end > How Jquery+ajax+springMVC returns json

How Jquery+ajax+springMVC returns json

一个新手Original: 2017-09-19 10:44:311336browse

JSP page:

<%@ page language="java" contentType="text/html; charset=utf-8" pageEncoding="utf-8"%>
<!DOCTYPE html>
<html>
<head>
	<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
	<title>登录验证</title>
  	<script type="text/javascript" src="js/jquery-1.8.3.js"></script>
  	<script type="text/javascript">
  		$(function(){  			
  			//ajax验证
  			$("#ajax1").click(function(){
  				alert(1);
  				$.post(
  					"ajax1.action",
  					function(result){
  						alert(2);
  						alert(result.ms);
  					}
  				);
  			});	
  		});
  	</script>
</head>

<body>
	<p id="con">
    	<p class="center"><h2>欢迎登录</h2></p>
    	<p>
    		<form action="login.action" method="post">
    		<table>
    			<tr>
    				<td>用户名：</td>
    				<td colspan="3"><input type="text" name="name"/></td>
    				<td></td>
    			</tr>
    			<tr>
    				<td>密 码：</td>
    				<td colspan="3"><input type="password" name="pwd"/></td>
    				<td></td>
    			</tr>
    			<tr>
    				<td>用户名：</td>
    				<td><input type="text" name="code"/></td>
    				<td></td>
    			</tr>
    			<tr>
    				<td class="center">
    					<input type="submit" name="sub" value="登录"/>
    				</td>
    			</tr>
    			<tr>
    				<td>
    					<button id="ajax1">ajax测试1</button>
    				</td>
    			</tr>
    		</table>
    		</form>
    	</p>
    </p>
</body>
</html>

Backend control class:

package com.login.control;

import javax.servlet.http.HttpServletRequest;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.ResponseBody;
import org.springframework.web.servlet.ModelAndView;

import com.alibaba.fastjson.JSONObject;
import com.login.bean.Users;
import com.login.service.UsersService;

/**
 * 用户控制类
 * @author
 *
 */
@Controller
public class UsersControl {
	
	@RequestMapping(value="ajax1.action",method= RequestMethod.POST)    	@ResponseBody
	public Object ajax1(HttpServletRequest request){
		System.out.println("进入控制类");
		JSONObject json = new JSONObject();
		String ms = "ajax测试1成功！哈哈!";
		json.put("ms", ms);
		System.out.println("出控制类");
		return json;
	}
}

You must add these two sentences to return json Value

@RequestMapping(value="ajax1.action",method= RequestMethod.POST)
@ResponseBody

The above is the detailed content of How Jquery+ajax+springMVC returns json. For more information, please follow other related articles on the PHP Chinese website!

Statement：

The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn

Previous article：Introduction to Foundation UI frameworkNext article：Introduction to Foundation UI framework

See more

How Jquery+ajax+springMVC returns json

Related articles