Servlet映射和HTTP状态404-ServletmappingandHTTPstatus404

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:

我的Servlet实现有问题。我使用Apache Tomcat作为Servlet引擎,使用Eclipse作为IDE。首先我创建了一个search.html如下:

    
    
    
    
    
        
        Name : 
        
        
    

Then I created a servlet, called Servlet1:

然后我创建了一个名为Servlet1的servlet:

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:

最后是位于WebContent \ WEB-INF中的web.xml:

  Servlet

  
    servlet1
    Servlet1
 
  
    servlet1
    /myServlet
  
  

When I enter data inside the input box of search.html and press "OK" I get the following error:

当我在search.html的输入框中输入数据并按“确定”时,我收到以下错误:

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.

我认为web.xml存在一些错误。事实上,如果我将url-pattern更改为es1,我不会收到错误。

1 个解决方案