Question

我正在使用Apache Tomcat版本制作一个简单的Web应用程序：8.0.36。我得到了＆＃34; Class foo.Counter不是Servlet＆＃34;错误。我在Servlet类中扩展了HttpServlet并配置了web.xml。我点击了极端的＆＃39;选择重复。

package foo;
import java.io.IOException;
import java.util.ArrayList;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class Counter extends HttpServlet{

@Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp)
        throws ServletException, IOException {

    String c=req.getParameter("hobby");

    Choices m=new Choices();
    ArrayList l=m.getNames(c);

    req.setAttribute("name", l);

    RequestDispatcher view=req.getRequestDispatcher("BasicCounter.jsp");
    view.forward(req, resp);
}
}

web.xml文件如下：

<?xml version="1.0" encoding="utf-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
version="2.5">

<servlet>
<servlet-name>Counter</servlet-name>
<servlet-class>foo.Counter</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>Counter</servlet-name>
<url-pattern>/HobbyParameter.do</url-pattern>
</servlet-mapping>

</web-app>

HTML文件为（names.html）：

<!DOCTYPE html>
<html>
<head>
<meta charset="ISO-8859-1">
<title>Insert title here</title>
</head>
<body>

<form method="POST" action="HobbyParameter.do">
Choose a Hobby:<br>

<select name="hobby" size="1">
<option>horse skiing
<option>extreme
<option>alpine scuba
<option>speed dating
</select>
<br>
<input type="SUBMIT" >

</form>

</body>
</html>

JSP as（BasicCounter.jsp）：

<%@page import="java.util.ArrayList"%>
<%@page import="java.util.Iterator"%>
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
The friends who share your hobby of <%= request.getAttribute("hobby") %> are:<br>
<%
ArrayList list= (ArrayList) request.getAttribute("name");
Iterator it=list.iterator();
while(it.hasNext())
{
out.print("<br>"+it.next());    
}
%>
</body>
</html>

错误如下：

SEVERE：为servlet Counter分配异常 java.lang.ClassCastException：foo.Counter无法强制转换为 javax.servlet.Servlet 在org.apache.catalina.core.StandardWrapper.loadServlet（StandardWrapper.java:1102）在 org.apache.catalina.core.StandardWrapper.allocate（StandardWrapper.java:828）在 org.apache.catalina.core.StandardWrapperValve.invoke（StandardWrapperValve.java:135）在org.apache.catalina.core.StandardContextValve.invoke（StandardContextValve.java:106）在org.apache.catalina.authenticator.AuthenticatorBase.invoke（AuthenticatorBase.java:502） at org.apache.catalina.core.StandardHostValve.invoke（StandardHostValve.java:141） at org.apache.catalina.valves.ErrorReportValve.invoke（ErrorReportValve.java:79）在org.apache.catalina.valves.AbstractAccessLogValve.invoke（AbstractAccessLogValve.java:616）在org.apache.catalina.core.StandardEngineValve.invoke（StandardEngineValve.java:88）在org.apache.catalina.connector.CoyoteAdapter.service（CoyoteAdapter.java:528）在org.apache.coyote.http11.AbstractHttp11Processor.process（AbstractHttp11Processor.java:1099） at org.apache.coyote.AbstractProtocol $ AbstractConnectionHandler.process（AbstractProtocol.java:670）在org.apache.tomcat.util.net.NioEndpoint $ SocketProcessor.doRun（NioEndpoint.java:1520）在org.apache.tomcat.util.net.NioEndpoint $ SocketProcessor.run（NioEndpoint.java:1476）在java.util.concurrent.ThreadPoolExecutor.runWorker（ThreadPoolExecutor.java:1142） at java.util.concurrent.ThreadPoolExecutor $ Worker.run（ThreadPoolExecutor.java:617） at org.apache.tomcat.util.threads.TaskThread $ WrappingRunnable.run（TaskThread.java:61）在java.lang.Thread.run（Thread.java:745）

Answer 1

在WEB-INF / lib中查找javax.servlet.ServletException.class的实例。 servler.jar或servlet.api.jar的内容应该由容器提供。所以你需要将servler.jar放入WEB-INF / lib文件夹并将jar添加到你的项目中。

Answer 2

看起来你已经为你的战争组装了servlet.jar。

如果您使用maven，则应将范围provided用于servlet api，如

<dependency>
 <groupId>javax.servlet</groupId>
 <artifactId>javax.servlet-api</artifactId>
 <version>3.0.1</version>
 <scope>provided</scope>
</dependency>

确保在Tomcat服务器中没有任何servlet jar依赖项，除了它自己的。

package.Class不是Servlet（Tomcat / J2EE）

2 个答案: