Has anyone found a solution for this? I'm trying the exact same thing and I'm having the exact same problem as @bhajan .
To recap: I'm using a standard SharePoint list with a PA custom form. If the user clicks on the 'New' button (in the standard SharePoint list, NOT in the custom form!! this is very important because some people here are giving wrong suggestions), I want the form to open on screen A. If the user clicks on the 'Edit' button in the standard list, I want the form to open on screen B. To avoid any confusion, I am talking about these two buttons in the SP list command bar.
These buttons cannot be referenced inside PowerApps as far as I know, so it's not possible to set an OnSelect or any other property.
In the App.StartScreen property I have the following code:
For debugging purposes I have added a label on the form that says true or false to indicate whether or not the if-condition is true.
However, no matter which case - true or false - the form always opens on screen A. I've been pulling my hair out over this.
I have tried all the variations you can think of, including IsBlank(SharePointIntegration.SelectedItemID) or IsEmpty(SharePointIntegration.Selected.Title) or whatever... the result remains the same.
I tried to work around this by putting a Navigate statement in scrA.OnVisible that would navigate to scrB if selected item was not blank, but apparently you're not allowed to use Navigate() in the OnVisible property. You cannot use Navigate() in the App.OnStart property, either so... I've no idea how else I can get this to work.
I feel like the above code should work and that this is a bug...
Oh btw, I have also set the OnNew and OnEdit properties of the SharePointIntegration connector to the correct screens, but this makes no difference whatsoever.
Any tips would be greatly appreciated because this really does look and feel like a bug to me...
