we are looking at building a inbound call flow with two possibilities based on two skill sets and availability.
How we will have it built is have skillets assigned to specific agents (3 for now) and if they are idle on queue to route to them via queue(Call this Skillet one),
and if they are not available we will route to a specific play message and route back to the same queue with another skill set (call this skill set two) where the rest of our agents will be on (around 20 agents).
the idea behind this is
Agents with skillet one, if a client calls they wont hear any IVR message
and agents with skillet two the client will hear a IVR message.
I assume this will be a switch statement but just need help with the logic of it.
The only thing you will need to do is decide when to play your message in the in-queue flow and make sure you only play it once. In this example, I assuming the timeout is 10 seconds. The 1st update data action creates a flow-level boolean variable Flow.alreadyPlayedMessage with the expression
This sets the value to false unless it's already been set; this is to keep each loop from resetting the value.
The decision action looks a Flow.alreadyPlayedMessage, and if it is false then play hold music for 10 seconds (or your timeout), then plays the message you want and sets Flow.alreadyPlayedMessage to true.
The hold music action at the bottom can be replaced with any other flow logic you want.
Notice if the decision action evaluates to true, that means your message has already been played once and the call will skip down to the hold music action at the end.
helped a lot, is there any chance to do this in the inbound call flow? when i implement above it detects that no flow has been played and routes to the queue i need and as expected but what we need is to have it for only a few agents and if no one available to straight away route to the rest.