SwiftUI List color background
@State var users: [String] = ["User 1", "User 2", "User 3", "User 4"]init(){ UITableView.appearance().backgroundColor = .red UITableViewCell.appearance().backgroundColor = .red UITableView.appearance().tableFooterView = UIView()}var body: some View { List(users, id: \.self){ user in Text(user) } .background(Color.red)}PlaygroundPage.current.liveView = UIHostingController(rootView: ContentView())
All SwiftUI's List
s are backed by a UITableView
in iOS. so you need to change the background color of the tableView. You make it clear
so other view
s will be visible underneath it:
struct ContentView: View { init(){ UITableView.appearance().backgroundColor = .clear } var body: some View { Form { Section(header: Text("First Section")) { Text("First cell") } Section(header: Text("Second Section")) { Text("First cell") } } .background(Color.yellow) }}
Now you can use Any background (including all Color
s) you want
Note that those top and bottom white areas are the safe areas and you can use the .edgesIgnoringSafeArea()
modifier to get rid of them.
Also note that List
with the .listStyle(GroupedListStyle())
modifier can be replaced by a simple Form
. But keep in mind that SwiftUI controls behave differently depending on their enclosing view.
Also you may want to set the UITableViewCell
's background color to clear
as well for plain tableviews
List
cannot be "painted". Please use the following instead:
List of items:
ForEach(items) { item in HStack { Text(item) }}.background(Color.red)
Scrollable list of items:
ScrollView { ForEach(items) { item in HStack { Text(item) } }.background(Color.red)}
In your case:
VStack { // or HStack for horizontal alignment Section(header: Text("Now in theaters")) { ScrollMovies(type: .currentMoviesInTheater) } Section(header: Text("Popular movies")) { ScrollMovies(type: .popularMovies) }}.background(Color.red)