You are viewing a plain text version of this content. The canonical link for it is here.
Posted to users@tomcat.apache.org by bharati <bh...@magnaquest.net> on 2009/12/10 06:38:22 UTC
URL Redirect in Tomcat 6
Hi,
Iam using Tomcat 6. In my applicaion, when Users entered any url like
'www.google.com' then it will check for the status of the User. If he is
Inactive, then it will send a http packet to redirect to the particular URL:
like "http://192.168.100.125:8080/" . Tomcat is listening to this port
8080.
My question is: how do I get the url from tomcat..? If manually user types
the url http://192.168.100.125:8080/ then it is displaying proper page to
login.
But in this case, Internet User does not know the URL. so...i need to get
the URL from Tomcat which is listening to 8080 Port.
Please help me.
Thanks in advance.
--
View this message in context: http://old.nabble.com/URL-Redirect-in-Tomcat-6-tp26722413p26722413.html
Sent from the Tomcat - User mailing list archive at Nabble.com.
---------------------------------------------------------------------
To unsubscribe, e-mail: users-unsubscribe@tomcat.apache.org
For additional commands, e-mail: users-help@tomcat.apache.org
RE: URL Redirect in Tomcat 6
Posted by "Hadole, Nishant IN BOM SISL" <Ni...@siemens.com>.
I guess she wants to display Customizable error responses, when application fails to respond (error code like 404 etc). Though, I agree, the question is not properly framed...
With best regards,
Nishant Hadole
Siemens IT Solutions and Services
SIS PRO SI-I
Tel.: +91 22 2495 7816
Fax: +91 22 6660 8521
Mailto: Nishant.Hadole@siemens.com
www.siemens.co.in
-----Original Message-----
From: Pid [mailto:pid@pidster.com]
Sent: Thursday, 10 December, 2009 02:29 PM
To: users@tomcat.apache.org
Subject: Re: URL Redirect in Tomcat 6
On 10/12/2009 05:38, bharati wrote:
>
> Hi,
> Iam using Tomcat 6. In my applicaion, when Users entered any url like
> 'www.google.com' then it will check for the status of the User.
>
> If he is Inactive, then it will send a http packet to redirect to the particular URL:
> like "http://192.168.100.125:8080/" . Tomcat is listening to this port
> 8080.
>
> My question is: how do I get the url from tomcat..? If manually user types
> the url http://192.168.100.125:8080/ then it is displaying proper page to
> login.
>
> But in this case, Internet User does not know the URL. so...i need to get
> the URL from Tomcat which is listening to 8080 Port.
>
> Please help me.
> Thanks in advance.
I'm baffled, I don't know about anyone else... Perhaps you could
rephrase your question in a way that might allow us to help you?
http://catb.org/~esr/faqs/smart-questions.html
p
---------------------------------------------------------------------
To unsubscribe, e-mail: users-unsubscribe@tomcat.apache.org
For additional commands, e-mail: users-help@tomcat.apache.org
Re: URL Redirect in Tomcat 6
Posted by Pid <pi...@pidster.com>.
On 10/12/2009 05:38, bharati wrote:
>
> Hi,
> Iam using Tomcat 6. In my applicaion, when Users entered any url like
> 'www.google.com' then it will check for the status of the User.
>
> If he is Inactive, then it will send a http packet to redirect to the particular URL:
> like "http://192.168.100.125:8080/" . Tomcat is listening to this port
> 8080.
>
> My question is: how do I get the url from tomcat..? If manually user types
> the url http://192.168.100.125:8080/ then it is displaying proper page to
> login.
>
> But in this case, Internet User does not know the URL. so...i need to get
> the URL from Tomcat which is listening to 8080 Port.
>
> Please help me.
> Thanks in advance.
I'm baffled, I don't know about anyone else... Perhaps you could
rephrase your question in a way that might allow us to help you?
http://catb.org/~esr/faqs/smart-questions.html
p
---------------------------------------------------------------------
To unsubscribe, e-mail: users-unsubscribe@tomcat.apache.org
For additional commands, e-mail: users-help@tomcat.apache.org
Re: URL Redirect in Tomcat 6
Posted by Christopher Schultz <ch...@christopherschultz.net>.
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
To whom it may concern,
On 12/10/2009 12:38 AM, bharati wrote:
> Iam using Tomcat 6. In my applicaion, when Users entered any url like
> 'www.google.com' then it will check for the status of the User.
Where does a user enter such a URL? Into the web browser's address bar?
If so, your application is not doing anything in this case (since
presumably the request will go to www.google.com).
> If he is
> Inactive, then it will send a http packet to redirect to the particular URL:
> like "http://192.168.100.125:8080/" . Tomcat is listening to this port
> 8080.
Okay, sounds good.
> My question is: how do I get the url from tomcat..?
Could it be as simple as request.getRequestURL()? This will return
"http://192.168.100.125:8080/" in the above example. If you're trying to
get the string "www.google.com", you'll probably have to try:
request.getHeader("Referer")
(note the intentional, historic misspelling of "referer")
This may be NULL if the client has been configured not to send the URL
of the referring page, though.
Is that what you were looking for? I agree with the other responses that
your question was quite unclear.
- -chris
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.4.10 (MingW32)
Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/
iEYEARECAAYFAkshfyIACgkQ9CaO5/Lv0PAJ8wCeKmg8Gj7/AIZb4FM5pWKyr0uI
fY8An1sUIwWWCvsGKZXH2kQUs+oe403w
=Qxrl
-----END PGP SIGNATURE-----
---------------------------------------------------------------------
To unsubscribe, e-mail: users-unsubscribe@tomcat.apache.org
For additional commands, e-mail: users-help@tomcat.apache.org