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我试图用非数组 @Published 项创建一个 ObservableObject。但是，我仍然不知道该怎么做。我尝试使用 ? 这样做。但是当我在视图中显示它时Text((details.info?.name)!)，它返回Thread 1: Swift runtime failure: force unwrapped a nil value我不知道问题是什么以及如何解决。我创建可观察对象类的方法是否正确？

class ShopDetailJSON: ObservableObject {
    
    @Published var info : Info?

    init(){load()}
    
    func load() {

        URLSession.shared.dataTask(with: request) { data, response, error in
            guard let data = data else {
                print("No data in response: \(error?.localizedDescription ?? "Unknown error").")
                return
            }
            if let decodedShopDetails = try? JSONDecoder().decode(ShopDetail.self, from: data) {
                DispatchQueue.main.async {
                    self.info = decodedShopDetails.info!
                }
            } else {
                print("Invalid response from server")
            }
        }.resume()
    }
    
}

struct Info : Codable, Identifiable {
    let contact : String?
    let name : String?
    var id = UUID()

    enum CodingKeys: String, CodingKey {

        case contact = "contact"
        case name = "name"
    }

    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        contact = try values.decodeIfPresent(String.self, forKey: .contact)
        name = try values.decodeIfPresent(String.self, forKey: .name)
    }

}

struct ShopDetail : Codable {
    let gallery : [Gallery]?
    let info : Info?

    enum CodingKeys: String, CodingKey {

        case gallery = "gallery"
        case info = "info"
    }

    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        gallery = try values.decodeIfPresent([Gallery].self, forKey: .gallery)
        info = try values.decodeIfPresent(Info.self, forKey: .info)
    }

}

示例 JSON 数据

{

    "gallery": [],
    "info": {
        "contact": "6012345678",
        "name": "My Salon",
    }
}
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1 回答 1

0

这是对您的代码中发生的事情的猜测，但是如果 JSON 数据从不为空，正如您在评论中所说，您可能正在尝试访问尚未更新的 ShopDetailJSON.info可选属性在你看来。

首先，进行一些清理工作。您不需要自定义实现init(from:)- 只要符合Codable您的情况就足够了。如果 JSON 值不是可选的，则无需将它们设为可选类型：

struct Info: Codable, Identifiable {
    let contact : String
    let name : String
    var id = UUID()
}

struct ShopDetail: Codable {
    let gallery : [Gallery]
    let info : Info
}

然后，当您获得 JSON 时，您不需要处理选项和强制展开!（无论如何都应该避免）：

if let decodedShopDetails = try? JSONDecoder().decode(ShopDetail.self, from: data {
    DispatchQueue.main.async {
        self.info = decodedShopDetails.info // no force-unwrap needed
    }
}

在视图中，您需要在访问其元素之前检查该info属性是否不存在。nil

struct ContentView: View {

   @ObservedObject var details: ShopDetailJSON

   var body: some View {
       Group() {

           // on iOS14
           if let info = details.info {
               Text(info.name)
           }

           // pre iOS14
           // if details.info != nil {
           //    Text(details.info!.name)
           // }
       }
       .onAppear {
           self.details.load()
       }
   }
}
于 2020-10-10T15:25:37.207 回答