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bdc604
Advocate III
Advocate III

Navigate to new item's display view after successful SubmitForm

Is there a way to navigate to the Display View of a newly created list item in SharePoint, something like:

 

OnSuccess of new form submit:

SharePointIntegration.SelectedListItemID=(FormNew1.LastSubmit.ID));Navigate(ScreenView, ScreenTransition.None)

1 ACCEPTED SOLUTION

Accepted Solutions
cjwallac
New Member

I was able to use SubmitForm(Form2_1);If(Form2_1.Valid, Navigate(MainScreen),""); to do something simuliar 

View solution in original post

3 REPLIES 3
v-micsh-msft
Community Support
Community Support

I don't think so.

 

SharePointIntegration.Selected should be ReadOnly, the selected item is passed through the SharePoint into the PowerApps integrated form under SharePoint, there is no available way to control that within PowerApps.

 

 

If you would like to show the created Item within the PowerApps form, you may take a try to add the SharePoint List from View->DataSource, then add a DisplayForm in a new screen, cofnigure the data source property to the SharePoint list, and Item property to:

FormNew1.LastSubmit

Which should be able to have the newly created item shown in the PowerApps form.

 

Regards,

Michael

Community Support Team _ Michael Shao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

I can set a global var with the value of the LastSubmit.ID:

 

Set(xLastID, FormNew1.LastSubmit.ID);
Navigate(ScreenView, ScreenTransition.None)

 

And potentially use it in the Item property as suggested.

However, formatting the funky powerapps lingo sucks:

 

If(xLastID="", SharePointIntegration.Selected, xLastID)

If(IsBlank(xLastID), SharePointIntegration.Selected, xLastID)

 

These don't work. It used to be so much easier to work with variables...

cjwallac
New Member

I was able to use SubmitForm(Form2_1);If(Form2_1.Valid, Navigate(MainScreen),""); to do something simuliar 

View solution in original post

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