SwiftUI 中的可选 @ObservableObject

Jon*_*gel 4 ios swift swiftui property-wrapper

我想在 SwiftUI 中有一个可选的 @ObservedObject 但我一直收到编译时错误。

Property type 'AModel?' does not match that of the 'wrappedValue' property of its wrapper type 'ObservedObject'
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这是一些最小的可重现代码。

import SwiftUI

public struct AView: View {
    
    //MARK: View Model
    //Error thrown here.
    @ObservedObject var model: AModel?
    
    //MARK: Body
    public var body: some View {
        Text("\(model?.value ?? 0)")
    }
    
    //MARK: Init
    public init() {
        
    }
    
}

class AModel: ObservableObject {
    let value: Int = 0
}
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New*_*Dev 5

技术原因是它Optional不是一个类,所以它不能符合ObservableObject.

但是如果你想避免重构对象本身——例如,如果它不在你的控制范围内——你可以这样做:

struct AView: View {

    private struct Content: View {
       @ObservedObject var model: AModel
       var body: some View {
          Text("\(model.value)")
       }
    }

    var model: AModel?
    
    var body: some View {
       if let model = model {
          Content(model: model)
       } else {
          Text("0")
       }
    }    
}
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