Answered

Using a collection to navigate through screens in a certain order

(0) Share

Report

Posted on by lalo1818

Using a collection for navigation.

I have a collection that gives me the order in which I want users to navigate through my app screens by hitting the next button.

The inital button click, goes to the first item/screen in the collection (screen 11 in the image). Then I want to the next button to navigate to the second item/screen in the collection (screen 6). As you hit the next button, I would like to navigate to the remainder of the screens in the collection: 15, 7, 5, 12, 3, 2, 10, 14....

The collection will shuffle each time so it needs to be dynamic. Any ideas? I appreciate any help.

Categories:

Building Power Apps

I have the same question (0)

All responses (4)

Answers (1)

WarrenBelz 153,075 Most Valuable Professional on at

Like (0)

Report
Hi @lalo1818 ,

Do a Collection like this (with your actual screen names)

ClearCollect( colRandomNum, Table( { Id: 11, Screen: Screen1 }, { Id: 6, Screen: Screen2 }, { Id: 15, Screen: Screen3 } ) )

And then

Navigate( LookUp( colRandomNum, Id = YourNumberChosen ).Screen )

Please click Accept as solution if my post helped you solve your issue. This will help others find it more readily. It also closes the item. If the content was useful in other ways, please consider giving it Thumbs Up.

MVP (Business Applications) Visit my blog Practical Power Apps

Was this reply helpful? Yes No
lalo1818 36 on at

Like (0)

Report

That is the way I had it, but then I added the shuffle function. So with the shuffle function, the screen names with be different evey time. Should I use a variable for the screen?

Was this reply helpful? Yes No
Verified answer

WarrenBelz 153,075 Most Valuable Professional on at

Like (1)

Report

Hi @lalo1818 ,

I do not believe what you want is possible (aligning random numbers with the correct screen - how do you choose the screen ?)

Was this reply helpful? Yes No
lalo1818 36 on at

Like (0)

Report

ok, good to know. I'll try different logic. I was using RandBetween, but it keeps giving me duplicates. Thank you for your help!

Was this reply helpful? Yes No