I have the following servlet code
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();
}
}
And the following stack trace
javax.servlet.ServletException: Class view.BackupServlet is not a 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)
root cause
java.lang.ClassCastException: view.BackupServlet cannot be cast to 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)
and the web.xml is
<?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>`
The servlet ErrorHandler is defined in usual way as any other servlet and configured in web. xml. If there is any error with status code either 404 (Not Found) or 403 (Forbidden ), then ErrorHandler servlet would be called.
To create a servlet, write a public Java class that includes basic I/O support as well as the package javax. servlet. The class must extend either GenericServlet or HttpServlet. Because Sun Java System Web Server 7.0 servlets exist in an HTTP environment, the latter class is recommended.
A servlet is a Java programming language class that is used to extend the capabilities of servers that host applications accessed by means of a request-response programming model. Although servlets can respond to any type of request, they are commonly used to extend the applications hosted by web servers.
For non-java developers, servlet is not suitable as they required to have a broad knowledge of Java servlet. HTML code is mixed up with Java code therefore, changes done in one code can affect another code. Writing HTML code in servlet programming is very difficult. It also makes servlet looks bulky.
You haven't shown your class declaration - my guess is that your class doesn't extend Servlet
or HttpServlet
.
If it does, then check how many different servlet.jar files you have in your deployment - it could be that it's being loaded by two different classloaders.
Your class containing the method doPost
must extend javax.servlet.Servlet
, but preferrably javax.servlet.HttpServlet
public class BackupServlet extends HttpServlet {
public void doPost(HttpServletRequest request,
HttpServletResponse response) {
.
.
.
}
}
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