你們好,今天鐵柱兄給你們帶一段jquery ajax提交數據給後端的教學。javascript
初學javaweb的同窗前端提交數據基本上都是用form表單提交,這玩意兒反正我是以爲不太好玩。而JavaScript ajax寫一大堆,看着都頭痛。jquery ajax簡單易懂容易學。html
廢話很少說,上教程~前端
新建一個Web項目,在\WebContent下新建一個index.jspjava
新建以後不用慌,默認的jsp編碼得改一下,我這邊統一改爲UTF-8:jquery
搞定以後咱們直接引入jquery的js文件,由於咱們村通網絡了,我就不想直接下載js了:web
直接引入js的網上路徑:<script src="https://cdn.staticfile.org/jquery/2.1.1/jquery.min.js"></script>ajax
簡簡單單,明明白白,寫兩個輸入框: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()">點我提交</a>//點擊事件
</body>
</html>
其實這裏我仍是想直接截圖的,可是懼怕大家噴我「啥做者,只會發圖片」。可是這裏確實沒啥好複製的。廢話很少說,我們繼續。後端
寫完這裏以後,先不急着寫js,我們先把後臺怎麼接收的給寫上。哈哈哈,又要發圖片了網絡
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類。我們回到jsp去,快快,跟上隊伍~
<%@ 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()">點我提交</a> <script type="text/javascript"> function btnConfirm(){//a標籤中的點擊事件 var userName=$("#userName").val();//經過id獲取輸入框中用戶輸入的值 var password=$("#password").val(); $.ajax({ type : 'post', url : '${pageContext.request.contextPath}/login', //這裏的/login對應LoginServlet類中註解的urlPatterns="/login" data:{'userName':userName,'password':password}, traditional : true, async : false, dataType: 'json', success : function(data){//成功的事件 alert("鐵柱兄真帥"); }, error : function(data){//失敗的事件 alert("你個衰仔!"); } }); } </script> </body> </html>
如今基本上就ok啦。ajax裏的各類動做我就不一一解說啦,百度裏面一大把哦。其實也不用知道是啥意思,能搞定用就行了。
如今咱們再去LoginServlet類裏去寫接收
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("接收到前端傳來的數據:userName="+userName+"password="+password); } }
這樣基本上沒啥毛病了,咱們把項目跑起來試一下
OK,後端能正常接收到前端傳來的值了。(那爲啥還說我是個衰仔?)
由於後端只接收了值,可是沒告訴ajax如今是啥狀況。咱們得返回點東西給ajax,告訴它咱們這邊一切正常。
resp.getWriter().write("666");隨便返回點東西給前端,只要有返回,ajax就知道你還活着了。
再跑一次~
好了。本次就到這裏啦。有什麼不懂的歡迎評論區討論~