andrewpaul
Programmer
I've been able to get PowerPrompts to do some pretty crazy stuff in the past but this latest request might be tad too far fetched.
Basically I need to have 1 page vs. several cascading pages. All the pull-down menus on the first page need to be related by way of their normal data relationships. That is, if I select one entity in one drop-down, then other entities that are not dependent on the first should dissapear and not be available via the other menus. I know this type of dynamic html is possible via Javascript as I have sample code that does it.
My theory is that I could dump all the data into Javascript arrays on the page. This would mean one big query or several smaller queries. At any rate, how would one get Oracle to come back with recordsets for 8 different pull-down boxes and you be able to make some semblence out of what data is a part of what?
On the database query end, I'm thinking that I would need n!/(n-2)! for n menus being related by pairs (representing n dynamos since each menu needs one). That comes out to 12 seperate combinations/permutations of any 2 menu dynamos (4!/(4-2)! = 12).
Any ideas on if this is the right way to do this?
Basically I need to have 1 page vs. several cascading pages. All the pull-down menus on the first page need to be related by way of their normal data relationships. That is, if I select one entity in one drop-down, then other entities that are not dependent on the first should dissapear and not be available via the other menus. I know this type of dynamic html is possible via Javascript as I have sample code that does it.
My theory is that I could dump all the data into Javascript arrays on the page. This would mean one big query or several smaller queries. At any rate, how would one get Oracle to come back with recordsets for 8 different pull-down boxes and you be able to make some semblence out of what data is a part of what?
On the database query end, I'm thinking that I would need n!/(n-2)! for n menus being related by pairs (representing n dynamos since each menu needs one). That comes out to 12 seperate combinations/permutations of any 2 menu dynamos (4!/(4-2)! = 12).
Any ideas on if this is the right way to do this?