Re: [WEB400] Tomcat app mapping

[Vanja Jovic <vanjajovic@xxxxxxxxxxxx>]

Aaron,

you need to add something like this in your web.xml file:

   <servlet>
     <servlet-name>CalcTax</servlet-name>
     <servlet-class>CalcTax</servlet-class>
     <init-param>
       <param-name>debug</param-name>
       <param-value>2</param-value>
     </init-param>
   </servlet>

...
...
   <servlet-mapping>
     <servlet-name>CalcTax</servlet-name>
     <url-pattern>/servlet/CalcTax</url-pattern>
   </servlet-mapping>


and you'll have to play a little with your server.xml file

Vanja Jovic,
Canada




ALBartell@taylorcorp.com wrote:

> I currently have an app running like so:
>
> http://172.111.111.111:8010/taylor/servlet/CalcTax
>
> but I would like to have the URL access be
>
> http://172.111.111.111:8010/servlet/CalcTax
>
> How do I change my <url-pattern> in <servlet-mapping> to do this?  Or
do I
> need to create a 'servlet' folder at the 'webapps' level and put my
serlvet
> in there?
>
> Let me know if I have not provided enough information,
> Aaron Bartell
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