java - 如何在 eclipse 和 tomcat 中创建简单的(servlet)java ee 项目

Question

我愿意

eclipse --> new --> dynamic web project --> in target runtime --> new Apache Tomcat v6.0 and establish mark Create a new local server --> next Browse establish way C:\dev\apache-tomcat-6.0.29-windows-x86\apache-tomcat-6.0.29 --> finish --> in new dynamic project finish

在 Java 资源中:src 新类

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 f extends HttpServlet 
{
    public void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException 
    {
        resp.setContentType("text/html;charset=utf-8");
        PrintWriter pw = resp.getWriter();
        pw.println("<H1>Hello, world! или Привет мир</H1>");
    }
}

-->server start in save resources s.java been modified. Save changes? I'm yes. --> Run servlet/s --> result HTTP Status 404 - /s/servlet/s

---

类型状态报告

消息/s/servlet/s

描述请求的资源(/s/servlet/s)不可用。

---

Apache Tomcat/6.0.14

我是如何以及何时出错的？怎么需要创建权限？

对不起，英语不好。解释得最清楚清楚

Answer 1

你检查过web.xml了吗？部署描述符？您必须在那里定义您的 servlet 映射，否则 servlet 容器将不会注意到该 servlet。

例子:

 <?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 definition -->
    <servlet>
        <servlet-name>s</servlet-name>
        <servlet-class>your.package.path.s</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

   <!-- servlet mapping -->
   <servlet-mapping>
       <servlet-name>s</servlet-name>
       <url-pattern>/myServlet</url-pattern>
   </servlet-mapping>
<web-app>

现在，您可以使用映射路径(“/<contextPath>/myServlet”)或默认调用程序(“/<contextPath>/servlet/s”)访问它。