September 1, 2017, 6:51pm
Sometimes I have to log another user in or out, and don’t have time to call them and tell them to do it themselves.
*45*xxx(their extension) doesn’t do it, it keeps logging me in and out instead.
Is there an easy way? If it involves running a bash script, it’s not worth it to me. Less time consumed asking them to do it.
September 5, 2017, 4:48pm
Ambiguous information. Expand this a little bit and help us help you. Logs from /var/log/asterisk/full for the call would help.
September 6, 2017, 7:07pm
From asterisk cli
To remove a queue member, you would use the queue remove member command:
*CLI> queue remove member SIP/0000FFFF0001 from support
Removed interface ‘SIP/0000FFFF0001’ from queue ‘support’
We built a replacement for iSymphony we use in house that utilizes login/logout via AMI.
but from your phone i just verified
This remotely logged out a user.
September 11, 2017, 4:33pm
*45(EXTEN) *(QUEUE) would almost be perfect but I can’t seem to do all queues at once with it
September 11, 2017, 6:07pm
That would be true. There is no mechanism in that particular feature to log someone out of all queues.
September 11, 2017, 6:15pm
Since asterisk doesn’t know of all your user queues a * feature code for log in/out of all queues is not possible without creating custom code.
You could write a web app that pulls a DB for all your user queues and members and then run AMI from your website to sign users in and out of the queues. Not sure how web dev you are but thats another way i am doing it.
I did not like iSymphony so we wrote our own.