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>123456789101112131415161718192021222324252627282930
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); } }123456789101112131415161718192021222324
运行效果:
作者:水巷石子
链接:https://blog.csdn.net/qq_37924905/article/details/108662526
来源:CSDN
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