首页 文章

我的类不是servlet错误

提问于
浏览 次
6

我有以下servlet代码

public void doPost(HttpServletRequest request, HttpServletResponse response){

    Backup bup = new Backup();
    bup.doBackup();

    response.setContentType("text/html");
    PrintWriter out;
    try {
        out = response.getWriter();

    out.println("You backed up your data - well hopefully");
    out.flush();
    out.close();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

}

以及下面的堆栈跟踪

javax.servlet.ServletException：类view.BackupServlet不是Servlet org.apache.catalina.valves.ErrorReportValve.invoke（ErrorReportValve.java:102）org.apache.catalina.connector.CoyoteAdapter.service（CoyoteAdapter.java:293 ）org.apache.coyote.http11.Http11Processor.process（Http11Processor.java:849)org.apache.coyote.http11.Http11Protocol $ Http11ConnectionHandler.process（Http11Protocol.java:583)org.apache.tomcat.util.net.JIoEndpoint $ Worker.run（JIoEndpoint.java:454）java.lang.Thread.run（Thread.java:619）根本原因java.lang.ClassCastException：view.BackupServlet无法强制转换为javax.servlet.Servlet org.apache.catalina .valves.ErrorReportValve.invoke（ErrorReportValve.java:102)org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:293)org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:849)org .apache.coyote.http11.Http11Protocol $ Http11ConnectionHandler.process（Http11Protocol.java:583)org.apache.tomcat.util.net.JIoEndpoint $ Worker.run（JIoEndpoint . java：454）java.lang.Thread.run（Thread.java:619）

和web.xml是

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-<br>app_2_4.xsd">
<display-name>Backup</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <description>BackupServlet</description>
    <display-name>BackupServlet</display-name>
    <servlet-name>BackupServlet</servlet-name><br>
    <servlet-class>view.BackupServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>BackupServlet</servlet-name>
    <url-pattern>/BackupServlet</url-pattern>
  </servlet-mapping>
</web-app>`
java servlets

3 回答

  • 13

    你没有't shown your class declaration - my guess is that your class doesn' t延伸 Servlet 或 HttpServlet .

    如果是，请检查部署中有多少个不同的servlet.jar文件 - 它可能是由两个不同的类加载器加载的 .

    回复于
  • 0

    包含方法 doPost 的类必须扩展 javax.servlet.Servlet ，但最好是 javax.servlet.HttpServlet

    public class BackupServlet extends HttpServlet {
    public void doPost(HttpServletRequest request,
                       HttpServletResponse response) {
     .
     .
     .
    }
}
    回复于
  • 3

    这很可能是您在服务器上加载了多个版本的servlet库（.jar文件） . 如果您使用maven / gradle作为构建工具，请确保将servlet库的范围更改为范围（如果它应由服务器提供） .

    回复于

相关问题