Link widget to specific UITableViewCell

Question

Created Jun ’20

Replies 4

Boosts 0

Views 1.2k

Participants 3

I followed along the Widget Code Along part two in my own app. In the tutorial they discussed linking the data on the widget to that in the tableview. When you clicked on the widget it would take you to the linked tableviewcell with swiftUI. In my app, my widget is swiftUI, but my tableview inside my main app is UIKit and Storyboards. How can create the URL link in UIKit like the one in swiftUI?

Answered by iEvanC in 624979022

Update
I figured out. I ended up using Universal URL, to open my app when the widget is clicked and go to the tableviewcell that correlates to the URL.

Boost

Answer 1

Developer Tools Engineer OP

Apple

Jun ’20

UIKit doesn’t have a special button or control specifically for links, but you can cause UIKit to open a URL by calling UIApplication.shared.open(<#url#>). For example, you might connect a UIButton to this IBAction:

Code Block swift@IBAction func openSupportSite(_ sender: Any) {
		UIApplication.shared.open(supportSiteURL)
}

Or put something like this in a table view delegate:

Code Block swiftfunc tableView(_: UITableView, didSelectRowAt indexPath: IndexPath) {
		let selectedRowData = rowData[indexPath.row]			// rowData is the array of objects you’re displaying
		UIApplication.shared.open(selectedRowData.url)
}

The method can also take a dictionary of options and a completion handler, but these parameters are optional.

0

Answer 2

iEvanC OP

Jul ’20

Not really what I'm looking for... I'm trying to use the widgetURL api to connect to my UITableView if that makes it clearer.

0

Answer 3

parimi1107 OP

Jul ’20

@iEvanC were you able to figure it out?

0

Answer 4

iEvanC OP

Aug ’20

Accepted Answer

Update
I figured out. I ended up using Universal URL, to open my app when the widget is clicked and go to the tableviewcell that correlates to the URL.

0