Hi It's me again.
I am sure about it. I think you want to point tomcat to some other directory then
webapps. (like when you open http://ADDRESS:PORT/, instead of tomcat page you
want to see your directory listing). To do that, create a new file called
ROOT.xml inside TOMCAT_HOME/conf/Catalina/localhost (default context directory).
Please cheak your tomcat documentation for specific context directory. ROOT.xml
contains:
<?xml version='1.0' encoding='utf-8'?>
<Context docBase="C:/Sample-dir" path="">
</Context>
That's it. Restart the tomcat and try http://ADDRESS:PORT/. You should see your
listing.
Hope I am right to understand your question this time. ;)
Regards,
Dhaval
--- Dhaval Patel <dhaval04@yahoo.com> wrote:
> Hi,
>
> Based upon your talk I am assuming following:
> - You have directory c:\sample-dir which text files and other files.
> - You want to make something like http://ADDRESS/sample-dir which displays list
> of files (like online apache browsing).
>
> Here is the one way you can do it quickly:
>
> (1) Even though your application is not web-app, you still need to make it look
> like web-app to tomcat. For that create directory "WEB-INF" inside
> c:\sample-dir.
> Now create a web.xml inside WEB-INF as following:
>
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <web-app xmlns="http://java.sun.com/xml/ns/j2ee"
> xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
> xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
> http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
> version="2.4">
> <servlet>
> <servlet-name>default</servlet-name>
>
> <servlet-class>org.apache.catalina.servlets.DefaultServlet</servlet-class>
> <init-param>
> <param-name>debug</param-name>
> <param-value>0</param-value>
> </init-param>
> <init-param>
> <param-name>listings</param-name>
> <param-value>true</param-value>
> </init-param>
> <load-on-startup>1</load-on-startup>
> </servlet>
> </web-app>
>
> I am running Tomcat 5.5 and I copied it from ROOT\WEB-INF\web.xml and I
> removed the contents between <web-app ..> and </web-app>. You can refer to
your
> tomcat installation. Basically all you need is the <web-app> node inside your
> web-xml. The <servlet> node is the default one. Make sure you set "listings" to
> true so that you can see the files. "listings" flag should be true by default.
> If
> it is true, you don't need <servlet> tag at all and <web-app> should be empty.
>
> (2) Now we have our sample-dir as a fake web application. Now we need to create
> context for it. There are two ways to do this:
>
> (A) Inside and detail way:
> Create the context file having name sample-dir.xml (why? because we want to
> call
> it as http://ADDRESS/samlple-dir. If you want to call it sample, make the
> filename sample.xml) ike this:
>
> <Context docBase="c:\sample-dir"
> privileged="true" antiResourceLocking="false" antiJARLocking="false">
> </Context>
>
> Where to put this xml? Depends upon your installation it should be inside
> conf. (probably conf\Catalina\localhost\). Just verify it and refer the
> documentation. It is very easy to find.
>
> (B) Easy way:
> Use "admin" webapplication of Tomcat to create context. It is very easy and
> it
> does above things for you.
>
> (3) Restart the tomcat on safe side and you should get the result on
> http://ADDRESS/sample-dir.
>
> Hope this helps. Let us know how it goes.
>
> Regards,
> Dhaval
>
> --- Smitha Murthy Krishna Nagesh <Smitha.Nagesh@Sun.COM> wrote:
>
> > I have a directory of text files that is not deployed into the Tomcat
> > server. It has some jsp's but certainly doesn't have the structure of a
> > web application. I need to somehow point Tomcat to this directory so
> > that the Tomcat can display the contents of these text files. Can you
> > please provide pointers on how to do it? Which files need to be modified
> > etc.
> >
> > Thanks,
> > Smitha
> >
> > --
> > Smitha Murthy Krishna Nagesh
> > Software RDE
> > Ph: (650) 786-6328
> >
> >
> >
> >
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> >
>
>
>
>
>
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