[Serusers] SER 0.9.6 "desc_time_order" issue

Mitya mitya at madein.hu
Mon Nov 12 19:34:57 CET 2007


Hi guys,

Does anybody have any comments on this ?

Thanks,
Mitya

Mitya wrote:
> Hi Everybody,
>
> I would like to set up thefollowing with ser 0.9.6
>
> 1)
> an UAC should be able to register more times with the same q value
>
> note: lets say, there is a power outage on the client side and it gets a 
> new IP from the ISP, registers again, the "old" registration has not 
> expired yed, it is still alive, because it did not unreggistered itself
>
> 2)
> when SER calls this UAC, only the most recently registered one should be 
> invited
>
> 3)
> to limit the maximum number of contacts is not possible (I need this 
> feature)
>
>
>  From the documentation:
> ------------------------
>
> ----- CUT ------
> 1.3.5. desc_time_order (integer)
>
> [...]
>
>    If set to 1 then all contacts will be ordered in descending
>    modification time order. In this case the most recently
>    updated/created contact will be used.
> [...]
>
>
> 1.3.3. append_branches (integer)
>
> [...]
>    If the parameter is set to 0, only
>    Request-URI will be overwritten with the highest-q rated
>    contact and the rest will be left unprocessed.
> [...]
> ----- CUT ------
>
>
>
> I also found this thread:
> -------------------------
> http://lists.iptel.org/pipermail/serusers/2003-March/000758.html
>
>  From Jan:
>
> You will have to use
> modparam("registrar", "append_branches", 0) in your script and after
> that only the most recently updated/created contact will be used (even if 
> there are many of them).
>
>
>
> This is exactly what I need !!
>
> my relevant config is the following:
>
> modparam("usrloc", "db_mode",   1)
> modparam("registrar", "append_branches", 0)
> modparam("registrar", "desc_time_order", 1)
> modparam("registrar", "max_contacts", 50)
>
>
> Sympthom:
> ---------
> When I call the UA, always the firstly registered one gets the INVITE, 
> and "desc_time_order" DOES NOT work :(
>
>
> Can sy explain me what should I do now ?
>
> Many thanks.
> Mitya
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>   




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