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From André Warnier>
Subject Re: Fwd: Is it not possible to install apache-tomcat-6.0.20 with a J2SE 5 JRE: A new development
Date Thu, 23 Jul 2009 12:28:14 GMT
Good, we are making progress !

Varuna Seneviratna wrote:
> Andre:
>      I edited the service.bat, only part that had to be added was "echo
> Using %EXECUTABLE%:"
> C:\apache-tomcat-6.0.20\bin>service.bat install
> Installing the service 'Tomcat6' ...
> Using CATALINA_HOME:    C:\apache-tomcat-6.0.20
> Using CATALINA_BASE:    C:\apache-tomcat-6.0.20
> Using JAVA_HOME:        C:\Program Files\Java\jdk1.6.0_14
> Using JVM:              C:\Program
> Files\Java\jdk1.6.0_14\jre\bin\server\jvm.dll
> Using C:\apache-tomcat-6.0.20\bin\tomcat6.exe:
> "C:\apache-tomcat-6.0.20\bin\tom
> cat6.exe" //IS//Tomcat6 --StartClass org.apache.catalina.startup.Bootstrap
> --Sto
> pClass org.apache.catalina.startup.Bootstrap --StartParams start
> --StopParams st
> op
> The service 'Tomcat6' has been installed.

Ah ! That's good, isn't it ?

> I copied and pasted the path to the "JAVA_HOME" may be that's what made the
> difference this time.

But if JAVA_HOME was pointed to the wrong path how
> could it be started in a non service mode.
Probably because then, you started Tomcat using the "startup.bat" 
command file, and not the parameters contained in the Registry.
It is certainly logical, but you would have to examine "startup.bat" to 
find the exact reason.
Do that as an exercise, it may help.

Now "apache tomcat6" is in the
> services list But can not be started.When tried to start
> the message is "Error 3: The system can not find the path specified".The
> "Using JVM" earlier was  C:\Program
> Files\Java\jdk1.6.0_14\jre\bin\client\jvm.dll and now it has become
> jre\bin\server\jvm.dll.In both these locations there is a jvm.dll
>                   In the Event Viewer under system the message was
> Event Type:    Error
> Event Source:    Service Control Manager
> Event Category:    None
> Event ID:    7000
> Date:        23/07/2009
> Time:        16:52:44
> User:        N/A
> Computer:    VARUNA
> Description:
> The Apache Tomcat 6 service failed to start due to the following error:
> The system cannot find the path specified.
Ok, so now we have only a path somewhere pointing to the wrong place.
But at least we know what the real problem is.

> For more information, see Help and Support Center at
Maybe you should try them ?  ;-)

> When given the long command
> C:\apache-tomcat-6.0.20\bin>"tomcat6.exe" //IS//%SERVICE_NAME% --StartClass
> org.
> apache.catalina.startup.Bootstrap --StopClass
> org.apache.catalina.startup.Bootst
> rap --StartParams start --StopParams stop
> The above command displayed a message box with a big Red X round shape
> mark(What do you call it?), titled "Application System"  and it's message
> was nonAlpha 37
That is a new one to me, I never saw one of those.
Maybe this is still some Kasperski thing ?

Anyway, now seems the appropriate time to see what our friend the 
"tomcat6w.exe" program has to say.  Remember, this is the GUI program 
that allows you to examine (and change) the Registry parameters that are 
used by tomcat6.exe when it tries to run the JVM and Tomcat.
Look at the various tabs there, and try to find something which does not 
We cannot really help you there, because of course this is only on your 
system that the parameters are wrong.  On our systems Tomcat starts up 
fine. ;-)

For example, on my Windows XP station, I have this under the "Java" tab :
Java Virtual Machine : C:\Java6jre\bin\client\jvm.dll
Java classpath : C:\Tomcat5.5\bin\bootstrap.jar
all of these paths should be correct of course.
What do yours look like ?

By the way, this "tomcat6w.exe" program is just a kind of "registry 
editor".  The values you see here are the ones that you can also see 
using the "regedit" program, and examining the key :
(but it is much safer doing this with tomcat6w.exe)

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