I think what you are looking for is to make your application the default
application for your Tomcat instance.  Search the archives about this,
since it is a common question. Basically you need to delete the existing
ROOT, and rename your app to be ROOT.

On Jul 31, 2008, at 10:50 AM, nilanthan wrote:


Yes I have a domain, ex, mydomain.com, I have as A-record going to that IP. The issue is that I want when someone types in mydomain.com in the address bar to forward to http://00.00.00.00:8080/mywebapp/welcome.action and still
display mydomain.com. I need the urlrewrite to forward mydomain.com to
http://00.00.00.00:8080/mywebapp/welcome.action but it wont accept the
mydomain.com to be forward, it only accepts anything that comes after
mydomain.com, for example, mydomain.com/mydomain.ca, the mydomain.ca gets rewritten only, not the mydomain.com part, which is what i want rewritten.


Ken Bowen wrote:

If you mean: type http://mydomain.com and get to 00.00.00.00:/,
you must register mydomain.com in the DNS with some ISP.
If you want to get rid of the :8080, you must configure Tomcat to
listen on port 80 instead of 8080 (check the howTo's, or search the
archives).

On Jul 30, 2008, at 1:13 PM, nilanthan wrote:


how do i  map to the domain name itself..

When I do /mydomain.com  to /mywebapp/welcome.action, it only works
if I
type this in the browser:
http://00.00.00.00:8080/mydomain.com.
then it goes to http://00.00.00.00:8080/mywebapp/welcome.action but
I have
created a record for mydomain.com to go to the ip of 00.00.00.00,
same as
above ip.

My question is how can I map to the mydomain.com to go to
http://00.00.00.00
and based on that go to http://00.00.00.00:8080/mywebapp/welcome.action?
?


nilanthan wrote:

Thanks Ken for all your help!


Ken Bowen wrote:

In rule, map

/mydomain.com  to /mywebapp/welcome.action

in outbound-rule, map

/mywebapp/welcome.action to  /mydomain.com

Do that for welcome, register, and every other page.

Depending on the rules you need, you can make some use of regular
expressions.

On Jul 29, 2008, at 10:25 AM, nilanthan wrote:


Thanks. I have placed the files/codes in the right spots and when
I do
localhost:8080/mywebapp/rewrite-status I get the urlrewrite page.

But what i need is when a user types in the browser
www.mydomain.com, it
should go to http://00.00.00.00:8080/mywebapp/welcome.action but
still
display http://www.mydomain.com and when I click on a link, it
should take
my to ex. http://00.00.00.00:8080/mywebapp/register.action but in
the
address bar show http://www.mydomain.com/register.action.

Can you use this example for the urlrewrite and where to place the
codes?

Thanks.



Ken Bowen wrote:

I don't know of any tutorial -- I found the documentation gave me
enough guidance.
It's really pretty straight-forward.
Drop a filter definition like this in your web.xml:

<filter>
  <filter-name>UrlRewriteFilter</filter-name>
  <filter-
class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</
filter-class>
<!--
  <init-param>
      <param-name>logLevel</param-name>
      <param-value>sysout:DEBUG</param-value>
  </init-param>
-->
</filter>

<filter-mapping>
  <filter-name>UrlRewriteFilter</filter-name>
  <url-pattern>/*</url-pattern>
  <dispatcher>FORWARD</dispatcher>
  <dispatcher>REQUEST</dispatcher>
</filter-mapping>

Then add a file urlrewrite.xml in your WEB-INF containing
mappings in
the following spirit:

<rule>
        <from>^/PrivacyPolicy$</from>
        <to type="forward">/PrivacyPolicy.do</to>
</rule>
<outbound-rule>
  <from>^/PrivacyPolicy.do$</from>
  <to>/PrivacyPolicy</to>
</outbound-rule>

The <outbound-rule> describes how to map something going from the
server to the browser,
and the (inbound) <rule> describes how to map what you mapped on
output (now coming back from the browser)
back into what you  need to see on input.

If you removed the comment symbols in the <filter> element, you
get
detailed debugging ouptut.

Hope this helps.
Ken

On Jul 28, 2008, at 6:16 PM, nilanthan wrote:


Thanks. I have looked at that before but am a bit confused
about the
instructions. Is there a good tutorial for this urlrewrite?


Ken Bowen wrote:

Apply a rewrite filter (http://tuckey.org/urlrewrite/) to map
the
ip
expression to what you want.

ken

On Jul 28, 2008, at 5:36 PM, nilanthan wrote:


So what Can I do so that that domain goes to that address but
shows
the
domain in the address bar?

Yuval Perlov wrote:

Where ever you forward, that's what the address bar shows


On Jul 28, 2008, at 7:45 PM, nilanthan wrote:


Hi,
I have a website hosted on netfirms. I have a domain,exmaple,
mydomain.com
and it forwards to an address http:/xx.xx.xxx.xx:8080/
folder1/
welcome.action
where xx is the ip of the server.

Im runningTomcat 5.5 alone without apache. The problem is
that
when
a users
goes to www.mydomain.com, it takes them to the site but in
the
address bar
it shows http:/xx.xx.xxx.xx:8080/folder1/welcome.action
instead of
mydomain.com.

Is this an issue with DNS or something in Tomcat? I will have
multiple sites
running in the future so I cannot place the site folder in
the
ROOT
directory.

Thanks.
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