无法访问Servlet

问题描述 投票:0回答:1

我有以下web.xml

<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app 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_3_0.xsd"
    version="3.0">

    <display-name>Example</display-name>

    <!-- Servlets -->
    <servlet>

        <servlet-name>OrderServlet</servlet-name>
        <servlet-class>OrderServlet</servlet-class>
    </servlet>

    <servlet-mapping>
        <servlet-name>OrderServlet</servlet-name>
        <url-pattern>/CreateOrder</url-pattern>
    </servlet-mapping>
    </web-app>

以下是我的ServletClass

package de.uniba.dsg.dsam.client;

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;


public class OrderServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;


    public OrderServlet() {
        super();

    }


    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

        request.getRequestDispatcher("/CreateOrder.jsp").forward(request, response);
    }


    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

        doGet(request, response);
    }

}

[现在,当我转到链接LocalHost:8080:/CreateOrder时,它不会访问它,只有当我访问它显示为HTTP Status 404 - Not Found的链接时,才完全没有错误。我已经筋疲力尽了,因为它应该很容易地运行,但是它不起作用,我尝试只用web.xml中定义的简单名称和jsp创建一个Servlet,但它不是可访问的,但是可以访问。

jsp jpa servlets
1个回答
0
投票

[servlet-class应该包含完整的类层次结构,包括程序包:

 <servlet-class>de.uniba.dsg.dsam.client.OrderServlet</servlet-class>
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