Difficulties with setting up Apache tomcat: 'Need a little help.

Greetings everyone,

I would appreciate a little help from all the gurus out there.....

Tomcat refuses to bring up my servlet. I know I've done something wrong somewhere but can't figure out where.

Now let me confess all my wrong deeds:

I'm trying to run servlets (and JSPs) by installing tomcat on j2sdk1.4.0 (and not using J2EE sdk).
I have apache Tomcat 4.1.18 installed.
I set the JAVA_HOME environment variable to the directory where the J2 SDK is installed on c: (c:\j2sdk1.4.0)
I also have APACHE_HOME set to the location of the tomcat installation. But then, while troubleshooting, I also added CATALINA_HOME and made it point to the same location. (My thinking is this wasn't nessesary, but it wouldn't hurt. Was I right?)
I have included all the environment variables below in case it would help....

In the webapps directory which resides in the tomcat installation directory, I have my directory named gsmsecurity, which contains a WEB-INF subdirectory. This WEB_INF subdirectory now contains to other subdirectories: lib and classes. My servlet resides inside the classes folder. Now the examples folder that comes with tomcat is on the same level with my gsmsecurity folder.

After starting tomcat, I point my browser to http://127.0.0.1:8080/examples/servlet/HelloWorldExample to bring up the HelloWorld example servlet. But when I do: http://127.0.0.1:8080/gsmsecurity/servlet/HelloServlet I get a tomcat error report page: HTTP status 404.
When I try different urls e.g http://127.0.0.1:8080/gsmsecurity/HelloServlet, the result is the same. Even when I put HelloWorldExample inside my classes folder and tried to bring it up, I didn't succed. My conclusion was that this was probably because there was no web.xml file in the WEB-INF folder. I put a minimal amount of things in the xml file and put it up, but the problem persists.

I would appreciate it greatly if someone could tell me what I'm doing wrong here.

Here's a listing of my Servlet. It's s simple one to let me know everything OK,before serious work begins

import java.io.PrintWriter;
import java.io.IOException;
import java.util.Enumeration;
import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
 
public class HelloServlet extends HttpServlet
{
    public void init(ServletConfig theConfiguration)
    {
    
    }
    
    public void doGet(HttpServletRequest theRequest,HttpServletResponse theResponse) throws IOException,ServletException
    {
        theResponse.setContentType("text.html");
PrintWriter aWriter = theResponse.getWriter();

aWriter.println("<HTML>");
aWriter.println("<HEAD>");
aWriter.println("<TITLE> Demo GSM Security Servlet Page </HTML>");
aWriter.println("</HEAD>");
aWriter.println("<BODY>");
aWriter.println("<H1> Demo Servlet <H1>");
aWriter.println("<P> Hello, and welcome to the GSM Security page<BR>");
aWriter.println("</BODY>");
aWriter.println("</HTML>");
aWriter.close();

}
}

A listing of my web.xml file:

<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.2//EN" 
"http://java.sun.com/j2se/dtds/web-app_2_2.dtd">
 
<web-app>

<display-name>Wireless Security application</display-name>
<description>
    This application uses a client - server model to implement a Wireless security mechanism for 
            GSM networks.
</description>
 

 
<context-param>
  <param-name>myParameter</param-name>
  <param-value>myValue</param-value>
      <description>
    This parameter represents nothing sensible......
          </description>
  </context-param>
 
<servlet>
   <servlet-name>HelloServlet</servlet-name>
   <description>
     This servlet does this and this.....  :) 
           </description>
   <servlet-class>HelloServlet</servlet-class>
   <init-param>
     <param-name>NoParameter</param-name>
             <param-value>NoValue</param-value>
    </init-param>
           <load-on-startup>5</load-on-startup>
</servlet>
        <session-config>
    
            
    <session-timeout>30</session-timeout>
</session-config>
 
</web-app>

his is the song tomcat sings anytime it starts up:

My environment variables:

ALLUSERSPROFILE=C:\Documents and Settings\All Users
APPDATA=C:\Documents and Settings\Administrator\Application Data
CATALINA_HOME=C:\Program Files\Apache Group\Tomcat 4.1
classpath=.;c:\j2sdk1.4.0\jre\lib\rt.jar;C:\Program Files\Apache Group\Tomcat 4.
1\common\lib\servlet.jar
CommonProgramFiles=C:\Program Files\Common Files
COMPUTERNAME=LAPTOPYINKA
ComSpec=C:\WINNT\system32\cmd.exe
HOMEDRIVE=C:
HOMEPATH=\
include=c:\program files\devstudio\vc\include;c:\program files\devstudio\vc\atl\
include;c:\program files\devstudio\vc\mfc\include;%include%
JAVA_HOME=c:\j2sdk1.4.0
lib=c:\program files\devstudio\vc\lib;c:\program files\devstudio\vc\mfc\lib;%lib
%
LOGONSERVER=\\LAPTOPYINKA
MSDevDir=C:\Program Files\DevStudio\SharedIDE
NUMBER_OF_PROCESSORS=1
OS=Windows_NT
Os2LibPath=C:\WINNT\system32\os2\dll;
Path=C:\WINNT\system32;C:\WINNT;C:\WINNT\System32\Wbem;.;c:\j2sdk1.4.0\bin;C:\Pr
ogram Files\Apache Group\Tomcat 4.1\bin;c:\program files\devstudio\sharedide\bin
\ide;c:\program files\devstudio\sharedide\bin;c:\program files\devstudio\vc\bin
PATHEXT=.COM;.EXE;.BAT;.CMD;.VBS;.VBE;.JS;.JSE;.WSF;.WSH
PROCESSOR_ARCHITECTURE=x86
PROCESSOR_IDENTIFIER=x86 Family 6 Model 8 Stepping 1, GenuineIntel
PROCESSOR_LEVEL=6
PROCESSOR_REVISION=0801
ProgramFiles=C:\Program Files
PROMPT=$P$G
SystemDrive=C:
SystemRoot=C:\WINNT
TEMP=C:\DOCUME~1\ADMINI~1\LOCALS~1\Temp
TMP=C:\DOCUME~1\ADMINI~1\LOCALS~1\Temp
TOMCAT_HOME=C:\Program Files\Apache Group\Tomcat 4.1
USERDOMAIN=LAPTOPYINKA
USERNAME=Administrator
USERPROFILE=C:\Documents and Settings\Administrator
windir=C:\WINNT

Another question I have is, why does http:\\localhost:8080\blah,blah not work for me?

Maybe it would also help if I mention that I am running Windows 2000 professional without IIS installed.

Thanks for reading up to this point. You could as well kindly pass a comment or two.......

Try adding a <servlet-mapping> entry in your web.xml file.
Also its a good idea, as well as good practice, to put your serlvet class in a package.

You appear to be using URLs with "/servlet/" in them. This gets you involved with the "invoker" servlet.
Read this FAQ here at the ranch and find out why this is a bad idea.

As Richard said - it is essential to put all classes involved in servlets in a package.
Bill

u can try this out...c if this works..

add the following in ur web.xml after <\servlet>

- <servlet-mapping>
<servlet-name>HelloServlet</servlet-name>
<url-pattern>/HelloServlet</url-pattern>
</servlet-mapping>

and then use the following url to access the servlet

http://127.0.0.1:8080/gsmsecurity/HelloServlet

Moving this thread to the more appropriate Apache/Tomcat forum.

Mark

Thanks so much for those answers.

Richard, I've put the class in a package and included the <servlet-mapping>, like you and Nidhi said I should.

And William, I read

The Long and Sordid Tale of the Invoker Servlet

you linked me to and I got the message.
I think I've got it now, only that the page that comes up is blank.
First, I did as the FAQ page said at the begining of the page:

In the file conf/web.xml find a section of text that looks like this:

[/code]

You need to uncomment the servlet-mapping and then restart Tomcat. Your application should work now....

I uncommented these lines.
I did this just to see the application work, at least. But knew something was wrong when instead of displaying a webpage, I got a text file downloaded. Then saw it was in the part where the servlet sends the header [code] theResponse.setContentType("text.html"); [/code] . I fixed that to read [code] theResponse.setContentType("text/html"); [/code]. Then I put the servlet in a package named mypackage ( I always prefer to keep things simple at first. The package structure could get deeper later), and I added the <servlet-mapping> tag. But all I get now is a blank page. What am I doing wrong?

In a bid to solve this, I included my classes directory in the classpath. Still didn't work. I then chaged the way output is sent to the browser in the servlet, like this
[code]
ServletOutputStream out = theResponse.getOutputStream();
out.println("<HTML>");
[/code]
I think it's something to do with my web.xml file.

Listing of my web.xml file:
[code]
<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.2//EN"
"http://java.sun.com/j2se/dtds/web-app_2_2.dtd">

<web-app>


<display-name>Wireless Security application</display-name>
<description>
This application uses a client - server model to implement a Wireless security mechanism for
GSM networks.
</description>



<context-param>
<param-name>myParameter</param-name>
<param-value>myValue</param-value>
<description>
This parameter represents nothing sensible......
</description>
</context-param>

<servlet>
<servlet-name>HelloServlet</servlet-name>
<description>
This servlet does this and this.....

</description>
<servlet-class>mypackage.HelloServlet</servlet-class>
<init-param>
<param-name>NoParameter</param-name>
<param-value>NoValue</param-value>
</init-param>
<load-on-startup>5</load-on-startup>
</servlet>

<servlet-mapping>
<servlet-name>HelloServlet</servlet-name>
<url-pattern>/gsm</url-pattern>
</servlet-mapping>

<session-config>


<session-timeout>30</session-timeout>
</session-config>

</web-app>
[/code]

Any useful suggestion would be greatly appreaciated.

Thanks.

Hi,

Your url to invoke the servlet must be like
http://localhost:8080/myapp/HelloServlet

Check u have the right folder structure like

ur web.xml must be some thing like

<servlet>
<servlet-name>HelloServlet</servlet-name>
<servlet-class>HelloServlet</servlet-class>
</servlet>

<servlet-mapping>
<servlet-name>HelloServlet</servlet-name>
<url-pattern>/HelloServlet</url-pattern>
</servlet-mapping>

and ur server.xml..verfiy u have added ur application
<Context path="/myapp" docBase="myapp" debug="7" reloadable="true">
</Context>

These are the things u need to run a servlet from ur context(ie an application like myapp)

and abt Invoker servlet..since now u have mapped the servlet,its no longer required.If the mapping is not done (prior to tomcat 4) ur URL to invoke servlet will be like http://localhost:8080/myapp/servlet/HelloServlet

i'm just putting all the points which i learned recently

Chandrasekhar S
[ September 29, 2004: Message edited by: Chandra Sekhar ]