Jquey框架的Ajax中的$.ajax请求方式
##$.ajax请求方式
什么是$.ajax请求方式
这种方式是将GET和POST方式合成一种请求
如何调用
$.ajax({键：值,键：值,键：值});
##代码案例：
jquery_ajxa_$ajax.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Title</title>
<script src="js/jquery-3.3.1.js"></script>
</head>
<script type="application/javascript">
function clickFn() {
$.ajax({
url:"s2",
async:true,
data:"username=bingbing&password=456",
type:"post",
dataType:"text",
success:function (data) {
alert(data)
},
error:function () {
alert("服务器发生了错误")
}
});
}
</script>
<body>
<!--设计一个按钮，一点击这个按钮，就向服务器发出异步请求-->
<input type="button" value="点我，发出ajax异步请求" onclick="clickFn()"/>
</body>
</html>
AjaxServlet2
package com.lbl.servlet;
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;
@WebServlet("/s2")
public class AjaxServlet2 extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
doGet(request, response);
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
System.out.println("收到了异步请求....s2");
response.getWriter().write("helloworld...s2");
String username = request.getParameter("username");
String password = request.getParameter("password");
System.out.println(username);
System.out.println(password);
}
}
运行效果：
