【问题标题】:Add a different modifier to each Action Sheet button为每个操作表按钮添加不同的修饰符
【发布时间】:2021-04-26 17:33:50
【问题描述】:

这可能是一个很简单的问题,但我找不到答案。

我正在尝试构建一个带有两个按钮(以及一个取消按钮)的 ActionSheet:

  1. “从图库中选择”按钮打开一个imagePicker,其中sourceType 设置为.photoLibrary
  2. “拍一张新照片”按钮打开一个imagePickersourceType 设置为.camera

我已经成功地制作了 ActionSheet 和 imagePicker,但是无法确定在哪里添加修饰符来告诉每个按钮应该使用哪个 sourceType。我设法将它添加到 ActionSheet 之外的 sheet() 修饰符中的普通按钮中,一切正常:

Button(action: {
                            self.show.toggle()
                        })
                        {Text("Take a new picture")}
                        .sheet(isPresented: self.$show, content: {
                            ImagePicker(sourceType: .camera, show: self.$show, image: self.$imageTemp)
                        })

但是,我看不到在 ActionSheet 中的何处包含此信息。非常感谢任何可以提供帮助的人,我希望这很清楚:-)

这是我的代码:

struct ContentView: View {

@State private var showingActionSheet = false
@State var imageTemp : Data = (UIImage(systemName: "photo.on.rectangle.angled")?.jpegData(compressionQuality: 1))!

var body: some View {
        NavigationView {
Image(uiImage: UIImage(data: imageTemp)!)
                        .onTapGesture {
                                        self.showingActionSheet = true
                                    }
                        .actionSheet(isPresented: $showingActionSheet) {
                            ActionSheet(title: Text("Image selector"), message: Text("Select an image"), buttons: [
                                .default(Text("Select from Gallery"))
                                    {
                                        self.show.toggle()
                                    },
                                .default(Text("Take new picture")) {
                                    self.show.toggle()
                                },
                                .cancel()
                            ]
                            )
                        }
                 }
         }
}

而且,以防万一,这是我的 imagePicker 的代码,尽管我认为这可能没有必要。

struct ImagePicker: UIViewControllerRepresentable {
    var sourceType: UIImagePickerController.SourceType = .photoLibrary
    
    @Binding var show: Bool
    @Binding var image: Data
    
    func makeCoordinator() -> ImagePicker.Coordinator {
        let imagePicker = UIImagePickerController()
        return ImagePicker.Coordinator(child1: self)
    }
    
    func makeUIViewController(context: UIViewControllerRepresentableContext<ImagePicker>) -> UIImagePickerController {
        let picker = UIImagePickerController()
        picker.delegate = context.coordinator
        picker.sourceType = sourceType
        return picker
    }

    func updateUIViewController(_ uiViewController: UIImagePickerController, context: UIViewControllerRepresentableContext<ImagePicker>) {

    }
    
    class Coordinator: NSObject, UIImagePickerControllerDelegate, UINavigationControllerDelegate {
        var child : ImagePicker
        init(child1: ImagePicker) {
            child = child1
        }
    
    
    func imagePickerControllerDidCancel(_ picker: UIImagePickerController) {
        self.child.show.toggle()
    }
    
    func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey: Any]) {
        let image = info[.originalImage]as! UIImage
        let data = image.jpegData(compressionQuality: 0.45)
        self.child.image = data!
        self.child.show.toggle()
    }
}
}

【问题讨论】:

    标签: swiftui imagepicker swiftui-actionsheet


    【解决方案1】:

    您的问题几乎可以归结为“我怎样才能呈现多张纸?”,所以this thread 可能会有所帮助。

    1. 定义一个新的enum 以包含可能的工作表类型(图库/拍照)
    2. 声明一个@State 属性来保存当前的工作表类型。它是可选的,因为当它为 nil 时,将不会显示任何工作表。
    3. 将属性设置为您想要的类型
    4. 使用sheet(item:onDismiss:content:) 而不是sheet(isPresented:onDismiss:content:)isPresented 最适合静态表。 item 适用于您有多种工作表类型时,这是您想要的。
    enum PhotoSheetType: Identifiable { /// 1.
        var id: UUID {
            UUID()
        }
        case gallery
        case picture
    }
    
    struct ContentView: View {
        
        /// 2.
        @State private var showingType: PhotoSheetType?
        @State private var showingActionSheet = false
        @State var imageTemp : Data = (UIImage(systemName: "photo.on.rectangle.angled")?.jpegData(compressionQuality: 1))!
        
        var body: some View {
            NavigationView {
                Image(uiImage: UIImage(data: imageTemp)!)
                    .onTapGesture {
                        self.showingActionSheet = true
                    }
                    .actionSheet(isPresented: $showingActionSheet) {
                        ActionSheet(
                            title: Text("Image selector"),
                            message: Text("Select an image"),
                            buttons: [
                                .default(Text("Select from Gallery")) {
                                    showingType = .gallery /// 3.
                                },
                                .default(Text("Take new picture")) {
                                    showingType = .picture /// 3.
                                },
                                .cancel()
                            ]
                        )
                    }            /// 4.
                    .sheet(item: $showingType) { type in 
                        if type == .gallery {
                            ImagePicker(sourceType: .photoLibrary, showingType: $showingType, image: self.$imageTemp)
                        } else {
                            ImagePicker(sourceType: .camera, showingType: $showingType, image: self.$imageTemp)
                        }
                    }
            }
        }
    }
    

    您还需要修改您的ImagePicker,以便Binding 接受PhotoSheetType? 而不是Bool。要关闭工作表,只需将 showingType 设置为 nil。

    struct ImagePicker: UIViewControllerRepresentable {
        var sourceType: UIImagePickerController.SourceType = .photoLibrary
        
        @Binding var showingType: PhotoSheetType?
        @Binding var image: Data
        
        func makeCoordinator() -> ImagePicker.Coordinator {
            let imagePicker = UIImagePickerController()
            return ImagePicker.Coordinator(child1: self)
        }
        
        func makeUIViewController(context: UIViewControllerRepresentableContext<ImagePicker>) -> UIImagePickerController {
            let picker = UIImagePickerController()
            picker.delegate = context.coordinator
            picker.sourceType = sourceType
            return picker
        }
        
        func updateUIViewController(_ uiViewController: UIImagePickerController, context: UIViewControllerRepresentableContext<ImagePicker>) {
            
        }
        
        class Coordinator: NSObject, UIImagePickerControllerDelegate, UINavigationControllerDelegate {
            var child : ImagePicker
            init(child1: ImagePicker) {
                child = child1
            }
            
            
            func imagePickerControllerDidCancel(_ picker: UIImagePickerController) {
                self.child.showingType = nil /// set to nil here
            }
            
            func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey: Any]) {
                let image = info[.originalImage] as! UIImage
                let data = image.jpegData(compressionQuality: 0.45)
                self.child.image = data!
                self.child.showingType = nil /// set to nil here
            }
        }
    }
    

    【讨论】:

    • 这是一个很棒的答案!超级清晰,解释清楚。你也是对的,它与另一个问题相似,我认为将代码放在 ActionSheet 中让我有点失望。唯一的事情是,当我尝试它时,我在控制台上收到消息“应用程序试图以模态方式呈现一个视图控制器 已经由 呈现”。看起来某个地方的“showingActionSheet”太多了,但我太菜鸟了,看不到确切的位置。
    • @FPL 上面的代码是否会出现消息?当我测试它很好时,问题可能在你的代码中的其他地方
    • 你说得对,我在别处发现了这个错误。非常感谢!
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