@WebServlet批注不适用于Tomcat 8

Question

我想@WebServlet在运行于Tomcat 8的Java EE Web应用程序中使用注释。

我已经读到我需要在自己的服务器中声明Servlet版本3.1，web.xml并且我的Servlet需要扩展HttpServlet。我做了所有这些，但仍然@WebServlet行不通。我正在获取HTTP 404。

我也试过我的配置用metadata-complete=”false”在我web.xml，但还是没有成功。

这是我的web.xml和Servlet。

完整的示例代码可以在GitHub上找到。

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app 
  version="3.1" 
  xmlns="http://xmlns.jcp.org/xml/ns/javaee" 
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
  xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">

  <context-param>
    <param-name>facelets.DEVELOPMENT</param-name>
    <param-value>true</param-value>
  </context-param>

  <!-- http://stackoverflow.com/a/7924117/451634 -->
  <!-- Put "-1" to disable this feature -->
  <context-param>
    <param-name>facelets.REFRESH_PERIOD</param-name>
    <param-value>1</param-value>
  </context-param>

  <servlet>
    <servlet-name>Faces Servlet</servlet-name>
    <servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>

  <servlet-mapping>
    <servlet-name>Faces Servlet</servlet-name>
    <url-pattern>*.xhtml</url-pattern>
  </servlet-mapping>

  <welcome-file-list>
    <welcome-file>index.xhtml</welcome-file>
  </welcome-file-list>

  <!-- JSF -->
  <listener>
    <listener-class>org.apache.myfaces.webapp.StartupServletContextListener</listener-class>
  </listener>   

  <!-- CDI -->
  <listener>
    <listener-class>org.apache.webbeans.servlet.WebBeansConfigurationListener</listener-class>
  </listener>

</web-app>

TestServlet.java

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

@WebServlet(name = "TestServlet", urlPatterns = {"*.serve"})
public class TestServlet extends HttpServlet {

  @Override
  protected void doGet(HttpServletRequest req, HttpServletResponse resp)
          throws ServletException, IOException {
    try (ServletOutputStream out = resp.getOutputStream()) {
      out.write("Hello World".getBytes());
      out.flush();
    }
  }

}

Answer 1

我知道了 我必须扩展启动Tomcat 8.0.12服务器的方式。

但是，必须完成三件事：

1. web-app version在web.xml中必须至少为3.0（我使用3.1）
2. metadata-complete在web.xml中可能不是真的（default is "false"）
3. classes 开始之前必须将目录添加到嵌入式Tomcat
   这是web.xml文件的示例：
   

<?xml version="1.0" encoding="UTF-8"?>
<web-app 
  version="3.1" 
  metadata-complete="false"  
  xmlns="http://xmlns.jcp.org/xml/ns/javaee" 
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
  xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">

  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
  </welcome-file-list>

</web-app>

这是我的Tomcat主类的样子（现在支持@WebServlet注释）：

public static void main(String[] args) throws Exception {
  String contextPath = "/";
  String webappDirLocation = "src/main/webapp/";
  String baseDirectory = new File(webappDirLocation).getAbsolutePath();

  Tomcat tomcat = new Tomcat();
  tomcat.setPort(8080);
  StandardContext context = (StandardContext) tomcat.addWebapp(contextPath, baseDirectory);

  // Additions to make @WebServlet work
  String buildPath = "target/classes";
  String webAppMount = "/WEB-INF/classes";

  File additionalWebInfClasses = new File(buildPath);
  WebResourceRoot resources = new StandardRoot(context);
  resources.addPreResources(new DirResourceSet(resources, webAppMount, additionalWebInfClasses.getAbsolutePath(), contextPath));
  context.setResources(resources);
  // End of additions

  tomcat.start();
  tomcat.getServer().await();
}