Re: [Linphone-developers] Android authorization username with '@' character

[Guillaume BERAUDO]

Hi Ahmed El-Refaey,


I am also surprised you succeeded in registering on iPhone but not on
Android.

You can check the information provided by your SIP provider with
linphone-desktop (a version exists for linux, mac and windows):
- create an account with the telephone number as username;
- on first attempt to connect you will be asked an authorization
  username and password;

On the linphone-mobile side, the "authorization username" field is
unfortunately not implemented yet (a unique username is used).


Keep me informed on the results of your tests.


Guillaume


Le Tue, 3 Jan 2012 20:31:59 +0200,
Ahmed Refaey <address@hidden> a écrit :

> Dear Community,
> 
> The User account I got from the SIP provider was as follows:
> Sip Account No.1
> 
> 
> > Display Name : Foo Bar
> > Telephone Number : 258963147852
> > Authorization User name :  address@hidden
> > Password : foobarpassword
> 
> 
> The interesting thing here which I got from the provider is that I
> use the authorization username as username(Tel. Number) which is not
> acceptable like this:
> 
> > 01-02 13:42:12.693: I/Linphone(361): From:
> > <sip:address@hidden
> > >;tag=835630168
> 
> and the provider stated that the successfull login message should be
> in that form:
> 
> > From: "Eduard Maya"
> > <sip:address@hidden:5061>;user=phone;
> 
> To: "Foo Bar" <sip:258963147852@ someserver.net:5060;user=phone >
> SIP/2.0
> 
> The question now is, how can I enter a username and Authorization
> username from the linphone/android interface which have only username
> field? and how when I entered the same authorization username in the
> iPhone client as username: "Foo%40" it worked successfully! but it
> did not work on the android?
> 
> Any help is highly appreciated,
> 
> *--*
> *Ahmed El-Refaey*
> 
> 
> 
> On Mon, Jan 2, 2012 at 1:59 PM, Ahmed Refaey <address@hidden>
> wrote:
> 
> > Dear Community,
> >
> > I was trying to register on SIP server which my
> > address@hidden, I tried this user name in the android
> > emulator, but it failed, I saw that the parsed username was only:
> > 'Ahmed' and the the full URI was: 'address@hidden@ xxx.com'
> >
> > I tried to replace the '@' character with '%40' and it failed also,
> > however this trick succeeded on iPhone client
> > here are the logs: http://pastebin.com/T55XeeAf
> >
> > Thanks in advance for fast reply,
> > *
> > *
> > *--*
> > *Ahmed El-Refaey*
> > *Senior Software Engineer*
> > ***R&D Team
> > *
> > *
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