Swift 将可解码结构转换为字典

编程问答 2022-05-30

问题描述

我创建了一个可解码的结构，用于解析存储在 firebase 中的数据。

struct discussionMessage: Decodable {
    var message: String
    var userCountryCode: String
    var userCountryEmoji: String
    var messageTimestamp: Double
    var userName: String
    var userEmailAddress: String
    var fcmToken: String?
    var question: String?
    var recordingUrl: String?
}

我想使用这个结构体也将数据存储在 firebase 中。但我收到错误：

*** 由于未捕获的异常“InvalidFirebaseData”而终止应用程序，原因：“(setValue:) 无法将 __SwiftValue 类型的对象存储在 .只能存储 NSNumber、Nsstring、NSDictionary 和 NSArray 类型的对象。 以 NSException 类型的未捕获异常终止

当我这样存储数据时：

let message = discussionMessage(message: messageTextView.text,userCountryCode: userCountryCode,userCountryEmoji: userCountryEmoji,messageTimestamp: timestamp,userName: userName,userEmailAddress: userEmail,fcmToken: nil,question: nil,recordingUrl: nil)       
        
messagesReference.childByAutoId().setValue(message)

有没有办法将可解码对象转换为字典，以便我可以将其存储在 firebase 中？

解决方法

你不需要它是可解码的。您需要的是能够对其进行编码（Encodable）。因此，首先将您的结构声明为 Codable。编码后，您可以使用 JSONSerialization jsonObject 方法将数据转换为字典：

extension Encodable {
    func data(using encoder: JSONEncoder = .init()) throws -> Data { try encoder.encode(self) }
    func string(using encoder: JSONEncoder = .init()) throws -> String { try data(using: encoder).string! }
    func dictionary(using encoder: JSONEncoder = .init(),options: JSONSerialization.ReadingOptions = []) throws -> [String: Any] {
        try JSONSerialization.jsonObject(with: try data(using: encoder),options: options) as? [String: Any] ?? [:]
    }
}

extension Data {
    func decodedObject<D: Decodable>(using decoder: JSONDecoder = .init()) throws -> D {
        try decoder.decode(D.self,from: self)
    }
}

extension Sequence where Element == UInt8 {
    var string: String? { String(bytes: self,encoding: .utf8) }
}

我还将 srtuct 属性声明为常量。如果您需要更改任何值，只需创建一个新对象：

struct DiscussionMessage: Codable {
    let message,userCountryCode,userCountryEmoji,userName,userEmailAddress: String
    let messageTimestamp: Double
    let fcmToken,question,recordingUrl: String?
}

let message: DiscussionMessage = .init(message: "message",userCountryCode: "BRA",userCountryEmoji: "??",userName: "userName",userEmailAddress: "email@address.com",messageTimestamp: 1610557474.227274,fcmToken: "fcmToken",question: "question",recordingUrl: nil)

do {
    let string = try message.string()
    print(string)      // {"fcmToken":"fcmToken","userName":"userName","message":"message","userCountryEmoji":"??","userEmailAddress":"email@address.com","question":"question","messageTimestamp":1610557474.2272739,"userCountryCode":"BRA"}
    
    let dictionary = try message.dictionary()
    print(dictionary)  // ["userName": userName,"userEmailAddress": email@address.com,"userCountryEmoji": ??,"messageTimestamp": 1610557474.227274,"question": question,"message": message,"fcmToken": fcmToken,"userCountryCode": BRA]
    
    let data = try message.data()      // 218 bytes
    let decodedMessages: DiscussionMessage = try data.decodedObject()
    print("decodedMessages",decodedMessages)    // ecodedMessages DiscussionMessage(message: "message",fcmToken: Optional("fcmToken"),question: Optional("question"),recordingUrl: nil)
} catch {
    print(error)
}
,

我要抛出一个答案，看看它是否有帮助。

我建议扩展 Encodable 以允许任何符合要求的对象返回可以写入 Firebase 的自身字典。

这是扩展程序

extension Encodable {
    var dict: [String: Any]? {
        guard let data = try? JSONEncoder().encode(self) else { return nil }
        return (try? JSONSerialization.jsonObject(with: data,options: .allowFragments)).flatMap { $0 as? [String: Any] }
    }
}

然后只是对结构的更改

struct DiscussionMessage: Encodable {
   ...your properties...
}

和代码

let msg = DiscussionMessage(data to populate with)
let dict = msg.dict
messagesReference.childByAutoId().setValue(dict)

请记住，实时数据库没有 nil - 任何没有值的节点都不存在，因此只会写入有值的属性。

哦 - 并确保属性是 NSNumber、NSString、NSDictionary 和 NSArray（和 Bool）的有效 Firebase 类型

decodablefirebasefirebasefirebasefirebase-realtime-databasestructstructstructswiftswiftswiftswift