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使用$.ajax方式实现页面异步访问,局部更新的效果

时间:2019-09-17 00:07:52      阅读:117      评论:0      收藏:0      [点我收藏+]
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Title</title>
<script src="js/jquery-3.3.1.min.js"></script>
<script>
function fun() {
$.ajax({
url:"ajaxServlet",
type:"POST",
data:{
"username":"light",
"age":12
},
success:function (data) {
alert(data);
},
error:function () {
alert("出错啦");
},
dataType:"text"
});
}

});
}

</script>
</head>
<body>
<input type="button" value="发送异步请求" onclick="fun();">
<input type="text">
</body>
</html>

package cn.hopetesting.com;

import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;

/**
* @author newcityman
* @date 2019/9/16 - 21:53
*/
@WebServlet("/ajaxServlet")
public class AjaxServlet extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String username = request.getParameter("username");
try {
Thread.sleep(5000);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println(username);
response.getWriter().write("hello,"+username+".");
}

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
this.doPost(request, response);
}
}
 

使用$.ajax方式实现页面异步访问,局部更新的效果

原文:https://www.cnblogs.com/newcityboy/p/11530830.html

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