I have a problem with my Servlet implementation. I'm using Apache Tomcat as the Servlet engine and Eclipse as the IDE. First I created a search.html as follow:
<html>
<head>
<meta charset="UTF-8">
<title>FirstServler</title>
</head>
<body>
<form action="/myServlet" method="get">
Name : <INPUT TYPE="text" NAME="name" SIZE="18"/>
<input type="submit" value="OK">
</form>
</body>
</html>
Then I created a servlet, called Servlet1:
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("/Servlet1")
public class Servlet1 extends HttpServlet
{
private static final long serialVersionUID = 1L;
public Servlet1()
{
super();
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
String s= request.getParameter("name");
response.getWriter().write(s);
}
}
And finally the web.xml, located in WebContent\WEB-INF:
<?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" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>Servlet</display-name>
<servlet>
<servlet-name>servlet1</servlet-name>
<servlet-class>Servlet1</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>servlet1</servlet-name>
<url-pattern>/myServlet</url-pattern>
</servlet-mapping>
</web-app>
When I enter data inside the input box of search.html and press "OK" I get the following error:
HTTP Status 404 - /Servlet1.0/es1
type Status report
message /Servlet1.0/es1
description The requested resource is not available.
I think there are some mistakes with web.xml. In fact, if I change the url-pattern into es1, I don't get an error.