Question

我试图使用ServletContextListener运行一个servlet，我已经从书中输入了代码 “Head-First”写作“Kathy sierra”，但这不起作用。它显示404错误。我已将类文件放在目录C：\ Tomcat 5.5 \ webapps \ Listener_exe \ web-inf \ classes \ com \ example 。和web-inf目录中的web.xml文件。所以请显示我的位置做错了。这是servlet，java文件和xml文件。

package com.example;

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

public class ListenerTester extends HttpServlet
{
public void doGet(HttpServletRequest request,HttpServletResponse response) throws IOException,ServletException
{
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("<html><body>test context attributes set by listener<br>");
Dog dog = (Dog) getServletContext().getAttribute("dog");
out.println("Dog's breed is: "+dog.getBreed()+</body></html>);
}
}

package com.example;

public class Dog
{
private String breed;
public Dog(String breed)
{
this.breed=breed;
}
public String getBreed()
{
return breed;
}
}

package com.example;

import javax.servlet.*;
public class MyServletContextListener implements ServletContextListener
{
public void contextInitialized(ServletContextEvent event)
{
ServletContext sc = event.getServletContext();
String dogBreed = sc.getInitParameter("breed");
Dog d = new Dog(dogBreed);
sc.setAttribute("dog",d);
}
public void contextDestroyed(ServletContextEvent event)
{}
}

<web-app>
<servlet>
<servlet-name>ListenerTester</servlet-name>
<servlet-class>com.example.ListenerTester</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ListenerTester</servlet-name>
<url-pattern>/ListenerTester</url-pattern>
</servlet-mapping>
<Context-param>
<param-name>breed</param-name>
<param-value>Great Dane</param-value>
</Context-param>
<listener>
<listener-class>
com.example.MyServletContextListener
</listener-class>
</listener>
</web-app>

Answer 1

您是否使用正确的网址来访问该网页？您的网址应该是，

http://localhost:8080/Listener_exe/ListenerTester

使用您为连接器设置的任何主机名或端口号。

Context Listener中的问题

1 个答案: