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ChanIsAvail - doesn't seem to workTechnical support, how-to guides, troubleshooting, and general assistance, from beginner to seasoned pro, this is where to discuss Asterisk, the most powerful open source PBX. |
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| Hi, I would like to be able to direct calls to the first phone in a group that is not already busy. It seemed to me that the ChanIsAvail application was the perfect way to do this - but it just doesn't seem to work as desired. So in the example below - call Sip/53 unless it is already on a call, then call Sip/52, unless it too is busy. exten => 66,1,ChanIsAvail(Sip/53&Sip/52|s) exten => 66,n,NoOp(${AVAILCHAN}) exten => 66,n,NoOp(${AVAILORIGCHAN}) exten => 66,n,NoOp(${AVAILSTATUS}) exten => 66,n,Dial(${AVAILORIGCHAN},${STDWAIT}) I am using the latest version of Asterisk - 1.4.11, and am testing with a mix of phones - Linksys SPA942, Grandstream GXP2000 & Snom 320. All yield similar results - regardless of whether the first phone is in a call or not, ChanIsAvail still returns ${AVAILCHAN} as the first phone in the list - in this case Sip/53. From the console :- -- Executing [66@internal:1] ChanIsAvail("SIP/50-170f5b00", "Sip/53&Sip/52|s") in new stack -- Executing [66@internal:2] NoOp("SIP/50-170f5b00", "SIP/53-170f9e30") in new stack -- Executing [66@internal:3] NoOp("SIP/50-170f5b00", "Sip/53") in new stack -- Executing [66@internal:4] NoOp("SIP/50-170f5b00", "0") in new stack -- Executing [66@internal:5] Dial("SIP/50-170f5b00", "Sip/53|20") in new stack Can anyone tell me how to get this working as desired? Thanks. |
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See ya... d.c. |
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| Thanks for that response D.C. I had seen and read through that particular article. I tried setting call-limit on the phone's sip.conf settings, it appeared to make no difference i.e. even when set to 1, and having a live call, then chanisavail still reported that phone as being available. Has anyone succeeded in getting a simple config working that sends a call to the first of a list of Sip phones that is not "in use"? I started to look into the "group" thing - but it seemed to get fairly complex fairly quick! It seems to me that many people must want to be able to do what I am trying to do here - someone out there has probably figured out how to do it. Please share the good news with us! Thanks. |
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