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.
>>>>>>>>> -- 
>>>>>>>>> View this message in context:
>>>>>>>>> http://www.nabble.com/address-bar-shows-ip-instead-of-domain-name-tp18694567p18694567.html
>>>>>>>>> Sent from the Tomcat - User mailing list archive at Nabble.com.
>>>>>>>>>
>>>>>>>>>
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>>>>>>>>>
>>>>>>>>
>>>>>>>>
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>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> -- 
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>>>>>>
>>>>>>
>>>>>>
>>>>>
>>>>> -- 
>>>>> View this message in context:
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>>>>> Sent from the Tomcat - User mailing list archive at Nabble.com.
>>>>>
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>>>>
>>>>
>>>>
>>>
>>> -- 
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>> 
>> 
>> 
> 
> 

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