Question

在Myeclipse中我创建了一个名为web1的web项目，并添加了一个名为servlet1的servlet，web.xml如下：

<?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>servlet1</servlet-name>
    <servlet-class>servlet1</servlet-class>
  </servlet>

  <servlet-mapping>
    <servlet-name>servlet1</servlet-name>
    <url-pattern>/test</url-pattern>
  </servlet-mapping>
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
</web-app>

但是当我在浏览器中输入地址：http：// localhost：8080 / web / test时，它没有用。我尝试了多次但没有回答。问题是什么？非常感谢！这是代码：

import java.io.IOException;
import java.io.PrintWriter;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;


public class servlet1 extends HttpServlet {

    /**
     * 
     */
    private static final long serialVersionUID = -6214906967399177511L;

    /**
     * Constructor of the object.
     */
    public servlet1() {
        super();
    }

    /**
     * Destruction of the servlet. <br>
     */
    public void destroy() {
        super.destroy(); // Just puts "destroy" string in log
        // Put your code here
    }

    /**
     * The doGet method of the servlet. <br>
     *
     * This method is called when a form has its tag value method equals to get.
     * 
     * @param request the request send by the client to the server
     * @param response the response send by the server to the client
     * @throws ServletException if an error occurred
     * @throws IOException if an error occurred
     */
    public void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {

        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        out.println("<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01 Transitional//EN\">");
        out.println("<HTML>");
        out.println("  <HEAD><TITLE>A Servlet</TITLE></HEAD>");
        out.println("  <BODY>");
        out.print("    This is ");
        out.print(this.getClass());
        out.println(", using the GET method");
        out.println("  </BODY>");
        out.println("</HTML>");
        out.flush();
        out.close();
    }

    /**
     * The doPost method of the servlet. <br>
     *
     * This method is called when a form has its tag value method equals to post.
     * 
     * @param request the request send by the client to the server
     * @param response the response send by the server to the client
     * @throws ServletException if an error occurred
     * @throws IOException if an error occurred
     */
    public void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {

        response.setContentType("text/html");
        PrintWriter out = response.getWriter();
        out.println("<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01 Transitional//EN\">");
        out.println("<HTML>");
        out.println("  <HEAD><TITLE>A Servlet</TITLE></HEAD>");
        out.println("  <BODY>");
        out.print("    This is ");
        out.print(this.getClass());
        out.println(", using the POST method");
        out.println("  </BODY>");
        out.println("</HTML>");
        out.flush();
        out.close();
    }

    /**
     * Initialization of the servlet. <br>
     *
     * @throws ServletException if an error occurs
     */
    public void init() throws ServletException {
        // Put your code here
    }

}

Answer 1

像往常一样，你忘了说super.init（config）;

 @Override
    public void init(ServletConfig config) throws ServletException {
        super.init(config);            
    }

你的servlet从未被初始化，因为你覆盖了方法，忘了打电话给父母＆＃39;初始化。

这就是为什么，如果你不了解内部，不要覆盖任何东西，除非你确定你在做什么。

servlet配置混乱

1 个答案: