Are you sure, that 'Demolog' is the fully qualified name of your Servlet? I
mean, don't you have a package name? (property <servlet-class>)
To load a Servlet at startup you need to provide some positive number as the
<load-on-startup> property.
If you want to call your Servlet from an URL like
http://localhost:8080/Demolog then you have to add a servlet-mapping to your
web.xml. Example
<servlet-mapping>
<servlet-name>Demolog</servlet-name>
<url-pattern>/Demolog</url-pattern>
</servlet-mapping>
hope this helps.
pero
> -----Original Message-----
> From: Jovie Castaneda [mailto:[EMAIL PROTECTED]]
> Sent: Monday, October 22, 2001 2:20 AM
> To: [EMAIL PROTECTED]
> Subject: RE: servlet in startup
>
>
> Im trying to load this servlet but I think im doing some funny thing here
> I edited the following tags in the web.xml
> <servlet-name>
> Demolog
> </servlet-name>
> <servlet-class>
> Demolog
> </servlet-class>
> <load-on-startup>
> -2147483646 ---> i am not sure of this so i retained this value
> </load-on-startup>
> ...then I tried to start Tomcat and it says that it "cannot load Demolog"
> How do I get around with this?
> Thanks in advance.
>
> -----Original Message-----
> From: Dmitri Colebatch [mailto:[EMAIL PROTECTED]]
> Sent: Monday, October 22, 2001 10:07 AM
> To: [EMAIL PROTECTED]
> Subject: Re: servlet in startup
>
>
> Look at the load on startup flag in web.xml
>
> hth
> dim
>
> On Mon, 22 Oct 2001, Jovie Castaneda wrote:
>
> > Hi guys,
> >
> > Can someone please help me how to make my servlet run when the Tomcat
> > starts up. What do I have to do in my codes
> > to impelement this or some setup needed for my Tomcat config.?
> > Thanks in advance
> >
> >
> >
>
>
