Question

..但我想我做的每件事都是正确的......请建议

我需要知道如何创建一个简单的servlet应用程序..在继续之前是否有任何设置在环境变量中的东西..我已经创建了helloworld servlet pgm但是获取404错误请求的资源不可用..

servlet类

package com.ignis;

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;

public class Hello extends HttpServlet {
  /**
     * 
     */
    private static final long serialVersionUID = 1L;

public void doGet(HttpServletRequest request,
                    HttpServletResponse response)
      throws ServletException, IOException {
    PrintWriter out = response.getWriter();
    out.println("Hello World");
  }
}

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>devProj</display-name>
    <welcome-file-list>
        <welcome-file>Hello</welcome-file>


    </welcome-file-list>

     <servlet>
        <servlet-name>HelloWorld</servlet-name>
        <servlet-class>Hello</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>HelloWorld</servlet-name>
        <url-pattern>/HelloWorld</url-pattern>
    </servlet-mapping>
</web-app>

Answer 1

servlet类声明应该是

 <servlet>
    <servlet-name>HelloWorld</servlet-name>
    <servlet-class>com.ignis.Hello</servlet-class>
</servlet>

您需要HttpServlet实施的完全限定类名。我很惊讶它甚至跑了。

Answer 2

您必须在web.xml中提供servlet类的完全限定名称。像这样：

<servlet>
    <servlet-name>HelloWorld</servlet-name>
    <servlet-class>com.ignis.Hello</servlet-class>
</servlet>

404错误，请求资源不可用

2 个答案: