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Full Version: [T46G] Action URI problem (SOLVED)
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Hello everybody

I try to call using the Action URI feature.
I'm not sure of the formating of the http request, please advise correct form.

FW version: 28.82.0.30

I tried:
Code:
http://IP_OF_MY_PHONE/servlet?p=login&q=loginForm&username=admin&pwd=MY_ADMIN_PASSWORD&jumpto=URI&key=number=NUMBER_I_WANT_TO_DIAL

http://IP_OF_MY_PHONE/servlet?p=login&q=login&username=admin&pwd=MY_ADMIN_PASSWORD&jumpto=URI&key=number=NUMBER_I_WANT_TO_DIAL

http://IP_OF_MY_PHONE/servlet?p=login&q=login&username=admin&pwd=MY_ADMIN_PASSWORD&jumpto=URI&key=number=NUMBER_I_WANT_TO_DIAL

The result is always the same, it opens a web page asking for my login credentials.

Thank you for your help.
Bump!
Hi,

Please try next string?
Code:
http(s)://username:password@<IP_OF_MY_PHONE>/servlet?key=number=NUMBER_I_WANT_TO_DIAL

Hope this will help.
Thanks for your input.

I found out that this format works in Firefox. Thanks a lot.
It is not working in IE though.

I consider this case closed for now.
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