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How to Verify Valid Menu Options in Shared Vector

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ssvec

Technical User
Jun 12, 2001
15
US
We would like to use a shared generic vector to play announcements and handle menu options for multiple VDN's. The VDN variables V1-V9 will be used to pass the announcement number and route to extensions.
For example, one of our menus has the announcement "press 1 for sales press 2 for service press 6 for assistance".
The VDN looks like:
VECTOR DIRECTORY NUMBER

VDN VARIABLES*

Var Description Assignment
V1 Announcement 1990010
V2
V3 Option 1 1061131111
V4 Option 2 1061132222
V5
V6
V7
V8 Option 6 1061133333
V9

The shared vector would be something like:

01 wait-time 2 secs hearing silence
02 collect 1 digits after announcement V1 for none
03 route-to number V3 with cov y if digit = 1
04 route-to number V4 with cov y if digit = 2
05 route-to number V5 with cov y if digit = 3
06 route-to number V6 with cov y if digit = 4
07 route-to number V7 with cov y if digit = 5
08 route-to number V8 with cov y if digit = 6
09 goto step 2 if unconditionally
10 stop
11
12


Our menus have variable numbers of options, and the options are often non-sequential. I have been unable to find a way to check that the option pressed was a valid option. In the above example, valid options would be 1,2, and 6. If the caller presses a non-valid option I would like to give them an announcement and then loop them back to step 2.

I played around with assigning the actual option numbers to V2, for example V2 = 126, and then tried to check in the vector if the digit pressed was "in" V2 but I am only able to do that using a table. If our options were sequential, like 0,1,2 I could have put the total options in V2 and then compared the digits to V2. We have too many existing menus with non-sequential options to make changing them impractical.

Any ideas would be greatly appreciated.
 
goto step x if V<1-9> < 1

If the variable is not set in the VDN it goes to a step to play an announcement stating the selection is invalid then you can drop, send back to top of menu, etc.
 
What is it doing now? I would assume if a caller presses 3 and V5 is not defined, the vector would fall through steps 6,7,8 and loop based on step 9. Is that not what is happenning?

-CL
 
Thanks for both of these responses! I didn't think about how it just might fall though. We are still in design phase so I have not tested it yet (my bad). I think I can make something work now.
 
User presses 5

goes past if digits = 1
goes past if digits = 2
etc..

goto step 20 if digits = 5

step 20 -> goto step 33 if V5 < 1
step 21 -> route to V5


step 33 -> announcement xxxx (sorry your selection is invalid)
step 34 -> goto step 1


 
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