你们好,今天铁柱兄给你们带一段jquery ajax提交数据给后端的教学。javascript
初学javaweb的同窗前端提交数据基本上都是用form表单提交,这玩意儿反正我是以为不太好玩。而JavaScript ajax写一大堆,看着都头痛。jquery ajax简单易懂容易学。html
废话很少说,上教程~前端
新建一个Web项目,在\WebContent下新建一个index.jspjava
新建以后不用慌,默认的jsp编码得改一下,我这边统一改为UTF-8:jquery
搞定以后咱们直接引入jquery的js文件,由于咱们村通网络了,我就不想直接下载js了:web
直接引入js的网上路径:<script src="https://cdn.staticfile.org/jquery/2.1.1/jquery.min.js"></script>ajax
简简单单,明明白白,写两个输入框:json
<%@ page language="java" contentType="text/html; charset=UTF-8"
pageEncoding="UTF-8"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<script src="https://cdn.staticfile.org/jquery/2.1.1/jquery.min.js"></script>
<title>Insert title here</title>
</head>
<body>
<input type="text" id="userName"/>
<input type="text" id="password"/>
<a onclick="btnConfirm()">点我提交</a>//点击事件
</body>
</html>
其实这里我仍是想直接截图的,可是惧怕大家喷我“啥做者,只会发图片”。可是这里确实没啥好复制的。废话很少说,我们继续。后端
写完这里以后,先不急着写js,我们先把后台怎么接收的给写上。哈哈哈,又要发图片了网络
package com.tiezhu.action; import java.io.IOException; import javax.servlet.ServletException; import javax.servlet.annotation.WebServlet; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; @WebServlet(name="LoginServlet",urlPatterns="/login") public class LoginServlet extends HttpServlet{ /** * */ private static final long serialVersionUID = 1L; @Override protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { // TODO Auto-generated method stub super.doGet(req, resp); } @Override protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { } }
好了,搞定java类。我们回到jsp去,快快,跟上队伍~
<%@ page language="java" contentType="text/html; charset=UTF-8" pageEncoding="UTF-8"%> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <script src="https://cdn.staticfile.org/jquery/2.1.1/jquery.min.js"></script> <title>Insert title here</title> </head> <body> <input type="text" id="userName"/> <input type="text" id="password"/> <a onclick="btnConfirm()">点我提交</a> <script type="text/javascript"> function btnConfirm(){//a标签中的点击事件 var userName=$("#userName").val();//经过id获取输入框中用户输入的值 var password=$("#password").val(); $.ajax({ type : 'post', url : '${pageContext.request.contextPath}/login', //这里的/login对应LoginServlet类中注解的urlPatterns="/login" data:{'userName':userName,'password':password}, traditional : true, async : false, dataType: 'json', success : function(data){//成功的事件 alert("铁柱兄真帅"); }, error : function(data){//失败的事件 alert("你个衰仔!"); } }); } </script> </body> </html>
如今基本上就ok啦。ajax里的各类动做我就不一一解说啦,百度里面一大把哦。其实也不用知道是啥意思,能搞定用就行了。
如今咱们再去LoginServlet类里去写接收
package com.tiezhu.action; import java.io.IOException; import javax.servlet.ServletException; import javax.servlet.annotation.WebServlet; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; @WebServlet(name="LoginServlet",urlPatterns="/login") public class LoginServlet extends HttpServlet{ /** * */ private static final long serialVersionUID = 1L; @Override protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { // TODO Auto-generated method stub super.doGet(req, resp); } @Override protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { String userName=req.getParameter("userName"); String password=req.getParameter("password"); System.out.println("接收到前端传来的数据:userName="+userName+"password="+password); } }
这样基本上没啥毛病了,咱们把项目跑起来试一下
OK,后端能正常接收到前端传来的值了。(那为啥还说我是个衰仔?)
由于后端只接收了值,可是没告诉ajax如今是啥状况。咱们得返回点东西给ajax,告诉它咱们这边一切正常。
resp.getWriter().write("666");随便返回点东西给前端,只要有返回,ajax就知道你还活着了。
再跑一次~
好了。本次就到这里啦。有什么不懂的欢迎评论区讨论~