iOS Foundations (SwiftUI) M4L9 challenge MenuPickerStyle()

[takipeti90]

Hello!

What can be the problem? MenuPickerStyle() doesn’t show the label “Location”, it shows the chosen location instead.

Here is the code:

HStack(spacing: 30) {
                
                Picker("Location", selection: $locationPickerIndex) {
                    
                    ForEach(0..<locations.count) { i in
                        Text(locations[i]).tag(i)
                    }
                    
                }
                .pickerStyle(MenuPickerStyle())
                
                Text(locations[locationPickerIndex])
                
            }

And here is the output of the simulator:

SegmentedPickerStyle() label works well.

[Chris_Parker]

What you are seeing is correct since you have chosen the picker style to be MenuPickerStyle(). The label is suppressed.

[takipeti90]

But in M4L9 SwiftUI Picker the MenuStylePicker shows the label “Tap Me” not the “Option 1…3”.

What is the difference here?

[Chris_Parker]

Hmmmmm, I suspect that this might have something to do with changes from Xcode 12 to Xcode 13 and moving from iOS 14 to iOS 15.

If I run the solution provided in the Course Resources I see exactly what you see and that’s what I also see in my own version of the challenge.

[Kyz]

Hi, did anyone find a solution to this problem? How to keep the “location” word displayed and only have the city change?

Thanks

[Abdul_Wahid]

I did a bit of a workaround by keeping my text label static as “Location” and making the picker as the label which shows the city

HStack {
                
                Text("Location:")
                
                
                Picker("Location", selection: $locationTag, content: {
                    
                    Text("Deira").tag(0)
                    Text("Nahda").tag(1)
                    Text("Tawar").tag(2)
                    Text("BurDubai").tag(3)
                    Text("Mamzar").tag(4)
                    Text("Muhaisna").tag(5)
                    
                    
                }).pickerStyle(MenuPickerStyle())
                
            }