Thank you Steve, and sorry for my "beginner's question".
But by me:

Here the line that start Tomcat in my startup.bat script:

call "%_TC_BIN_DIR%\tomcat" start %1 %2 %3 %4 %5 %6 %7 %8 %9

which launch the tomcat.bat script with the parameter "start".

You speak about a java command but it seems it is in the tomcat.bat script.

Should I modify the tomcat.bat?
I would prefer to modify the startup.bat! But what is the command in this case?

I have also tried to add a "CATALINA_OPTS" environment variable with the value 
"-Dhttp.proxyHost=http://proxy.blabla.com -Dhttp.proxyPort=8080" but that doesn't 
work. Maybe a proxy access problem?!?

Thank you very much
Sylvain


-----Message d'origine-----
De: Steven Noels [mailto:[EMAIL PROTECTED]]
Date: mardi, 8. octobre 2002 20:28
À: [EMAIL PROTECTED]
Objet: Re: File generator problem (fetch from an URI)


[EMAIL PROTECTED] wrote:

> what do you mean about "the java command" in the tomcat startup script?

I assume you run Tomcat using some startup script (batch file), 
%CATALINA_HOME%\bin\startup.bat comes with the distribution. That script 
starts a Java application, where you can set an HTTP proxy using 
commandline parameters (-Dhttp.proxyPort etc...)

They made it easy to add those parameters since you can also store them 
in an environment variable called CATALINA_OPTS, which is added to the 
line in the %CATALINA_HOME%\bin\startup.bat that starts Tomcat:

%_STARTJAVA% %CATALINA_OPTS% -Dcatalina.base="%CATALINA_BASE%" 
-Dcatalina.home="%CATALINA_HOME%" org.apache.catalina.startup.Bootstrap 
%2 %3 %4 %5 %6 %7 %8 %9 start

OK now?

</Steven>
-- 
Steven Noels                            http://outerthought.org/
Outerthought - Open Source, Java & XML Competence Support Center
[EMAIL PROTECTED]                      [EMAIL PROTECTED]


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