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我是初学者到servlet ...我想从html传递一个简单的参数到servlet，但是错误“HTTP Status 404 - Not Found”applears ..这里是我的代码... web .XML servalet java错误
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<servlet>
<servlet-name>NewClass</servlet-name>
<servlet-class>NewClass</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>NewClass</servlet-name>
<url-pattern>/NewClass</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>newhtml.html</welcome-file>
</welcome-file-list>
</web-app>newhtml.xml
<html>
<head>
<title>Simple Application</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<form action="/NewClass" >
<input name="username" />
<input type="submit"/>
</form>
</body>
</html><html>
<head>
<title>Simple Application</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<form action="/NewClass" >
<input name="username" />
<input type="submit"/>
</form>
</body>
</html><html>
<head>
<title>Simple Application</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
</head>
<body>
<form action="/NewClass" >
<input name="username" />
<input type="submit"/>
</form>
</body>
</html>NewClass.java
public class NewClass extends HttpServlet{
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException{
PrintWriter out = response.getWriter();
String req = request.getParameter("username");
out.println(req);}}请有人告诉我为什么一个错误出现在我身上？
你是如何将你的webapp部署到tomcat的，你的webapp的名字是什么？ – folkol 2014-09-20 14:16:55
您可能没有连接到服务器。如果在IDE之外启动，请确保启动服务器。另外，确保你输入正确的地址，http：// localhost：8080 /'。如果你正在做一些教程，那么请确保你按照这封信的每一步。 – MarGar 2014-09-20 14:17:02
他确实得到了404，这在没有服务器的情况下有些困难：P – folkol 2014-09-20 14:17:43