4
我想從jQuery的.ajax()函數調用一個servlet。
此刻,我不認爲我甚至要求servlet或傳遞PARAMATERS到它,但是很多谷歌搜索似乎並沒有幫助。有任何想法嗎?
這是我的html:
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
function login(){
$("#loading").hide();
var email = document.nameForm.email.value;
$.ajax({
type: "GET",
url: "ProcessForm",
data: "email="+email,
success: function(result){
alert(result);
}
});
}
</script>
<title>My AJAX</title>
</head>
<body>
<p>This time it's gonna work</p>
<form name="nameForm" id="nameForm" method="post" action="javascript:login()">
電子郵件 裝載
</body>
</html>
而且我的web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>ajaxtry</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>ProcessForm</servlet-name>
<servlet-class>com.ajaxtry.web.ProcesFormServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ProcessForm</servlet-name>
<url-pattern>/ProcessForm</url-pattern>
</servlet-mapping>
</web-app>
該servlet只是此刻模板:
package com.ajaxtry.web;
// imports here
public class ProcessFormServlet {
public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
System.out.println(request.getParameter("email"));
}
}
謝謝,我已經完成了前兩件事,第二件事是一個很好的建議,我認爲這將是一條路。 – Ankur 2009-12-02 04:11:47
我的意思是第三個,第二個是一個明顯的錯誤;) – Ankur 2009-12-02 04:42:12