Jaer_zk通过本文主要向大家介绍了jQuery Ajax前后端使用JSON进行交互示例等相关知识,希望对您有所帮助,也希望大家支持linkedu.com www.linkedu.com
需求：
前端通过jQuery Ajax传输json到后端，后端接收json，对json进行处理，后端返回一个json给前端
这里使用servlet的方式
1、采用$.post方法
index.jsp页面
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<%@ page contentType="text/html; charset=UTF-8"%>
<html>
<head>
<title></title>
<script src="js/jquery-1.12.2.js"></script>
<script language="JavaScript">
function checkUserid() {
$.post('Ajax/CheckServlet',//url
{
userid : $("#userid").val(),
sex : "男"
}, function(data) {
var obj = eval('(' + data + ')');
alert(obj.success);
});
}
</script>
</head>
<body>
用户ID:
<input type="text" id="userid" name="userid"> <span id="msg"></span>
<br> <button onclick="checkUserid()">传输</button>
</body>
</html>
</div>
CheckServlet.Java代码如下
package com.ajax;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class CheckServlet extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
this.doPost(request, response);
}
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
/*设置字符集为'UTF-8'*/
request.setCharacterEncoding("UTF-8");
response.setCharacterEncoding("UTF-8");
String userid = request.getParameter("userid"); // 接收userid
String sex = request.getParameter("sex");//接收性别
System.out.println(userid);
System.out.println(sex);
//写返回的JSON
PrintWriter pw = response.getWriter();
String json = "{'success':'成功','false':'失败'}";
pw.print(json);
pw.flush();
pw.close();
}
}
</div>
由于这里采用的是servlet的方式，所以要配置web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
id="WebApp_ID" version="3.1">
<display-name>Ajax</display-name>
<servlet>
<servlet-name>CheckServlet</servlet-name>
<servlet-class>com.ajax.CheckServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>CheckServlet</servlet-name>
<url-pattern>/Ajax/CheckServlet</url-pattern>
</servlet-mapping>
</web-app>
</div>
在页面输入一个ID，可以在后台接收到并且打印出来，后台通过PrintWriter进行回写JSON返回前端，前端通过eval将JSON变换为Object对象，通过obj.name获取JSON值
2、采用$.get方法，只需要将jsp页面里面的post改为get即可
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<%@ page contentType="text/html; charset=UTF-8"%>
<html>
<head>
<title></title>
<script src="js/jquery-1.12.2.js"></script>
<script language="JavaScript">
function checkUserid() {
$.get(
'Ajax/CheckServlet',//url
{
userid:$("#userid").val(),
sex:"男"
},
function(data){
var obj = eval('('+data+')');
alert(obj.success);
}
);
}
</script>
</head>
<body>
用户ID:
<input type="text" id="userid" name="userid"> <span id="msg"></span>
<br>
<button onclick="checkUserid()">传输</button>
</body>
</html>
</div>
结果与$.post一样
3、通过$.ajax方法
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<%@ page contentType="text/html; charset=UTF-8"%>
<html>
<head>
<title></title>
<script src="js/jquery-1.12.2.js"></script>
<script language="JavaScript">
function checkUserid() {
$.ajax({
type : 'post',
data : {
userid : $("#userid").val(),
sex : "男"
},
url : "Ajax/CheckServlet",
success : function(data) {
var obj = eval('(' + data + ')');
alert(obj.success);
},
error : function() {
},
complete : function() {
}
});
}
</script>
</head>
<body>
用户ID:
<input type="text" id="userid" name="userid"> <span id="msg"></span>
<br>
<button onclick="checkUserid()">传输</button>
</body>
</html>
</div>
$.ajax方法也是可以分为post和get方法的，通过修改type来修改发送的方式
结果与方法1是相同的
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。
</div>