Java Web应用程序指定入口点 [英] Java Web Application specify entry point

查看:484
本文介绍了Java Web应用程序指定入口点的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有一些Java Web应用程序,现在它从index.jsp页面开始. 我有自己的课程,并附有以下代码:

I have some Java Web Application and now it's starting from index.jsp page. I have my own class with following code:

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

public class EntryPointClass extends HttpServlet {
    @Override
    public void doGet(HttpServletRequest request, HttpServletResponse response)
    throws IOException, ServletException
    {
      ....
    }
}

我的MANIFEST.MF文件如下:

My MANIFEST.MF file looks like:

Manifest-Version: 1.0
Main-class: asap.vito.mobileapi.EntryPointClass

但是无论如何,删除index.jsp后,我遇到了404错误.

But anyway, after deleting index.jsp I have a 404 error.

要直接从我的java类启动应用程序,我应该怎么做?

What should I change to start my app directly from my java class?

谢谢.

推荐答案

首先是一个servlet,而不是简单的java类... 您应该将其映射到web.xml中,然后您可以直接访问它

First of all its a servlet not simple java class... You should give its mapping into the web.xml and then u can access it directly

为您的servlet提供这种类型的映射

give this type of mapping for your servlet

<welcome-file-list>
  <welcome-file>entryPointClass </welcome-file>
 </welcome-file-list>
    <servlet>
            <servlet-name>EntryPointClass </servlet-name>//servlet name here
            <servlet-class>examples.EntryPointClass </servlet-class> // packagename.servletclassname
        </servlet>

        <servlet-mapping>
            <servlet-name>EntryPointClass </servlet-name>// servletname
            <url-pattern>/entryPointClass </url-pattern>  // url parttern by which you will access your servlet
        </servlet-mapping>

要从服务器访问此文件,请使用projectname/urlpattern

To access this from server use projectname/urlpattern

这篇关于Java Web应用程序指定入口点的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆