正确解析 Swift 中的 JSON 3

Question

我正在尝试获取 JSON 响应并将结果存储在一个变量中。在 Xcode 8 的 GM 版本发布之前，我已经在之前的 Swift 版本中使用了此代码的版本。我在 Whosebug 上查看了一些类似的帖子： and 。

但是，这里传达的想法似乎不适用于这种情况。

如何正确解析 Swift 3 中的 JSON 响应？在 Swift 3 中读取 JSON 的方式有什么变化吗？

下面是有问题的代码（在 playground 中可以是运行）：

import Cocoa

let url = "https://api.forecast.io/forecast/apiKey/37.5673776,122.048951"

if let url = NSURL(string: url) {
    if let data = try? Data(contentsOf: url as URL) {
        do {
            let parsedData = try JSONSerialization.jsonObject(with: data as Data, options: .allowFragments)

        //Store response in NSDictionary for easy access
        let dict = parsedData as? NSDictionary

        let currentConditions = "\(dict!["currently"]!)"

        //This produces an error, Type 'Any' has no subscript members
        let currentTemperatureF = ("\(dict!["currently"]!["temperature"]!!)" as NSString).doubleValue

            //Display all current conditions from API
            print(currentConditions)

            //Output the current temperature in Fahrenheit
            print(currentTemperatureF)

        }
        //else throw an error detailing what went wrong
        catch let error as NSError {
            print("Details of JSON parsing error:\n \(error)")
        }
    }
}

编辑： 这是 print(currentConditions)

之后 API 调用的结果示例

["icon": partly-cloudy-night, "precipProbability": 0, "pressure": 1015.39, "humidity": 0.75, "precipIntensity": 0, "windSpeed": 6.04, "summary": Partly Cloudy, "ozone": 321.13, "temperature": 49.45, "dewPoint": 41.75, "apparentTemperature": 47, "windBearing": 332, "cloudCover": 0.28, "time": 1480846460]

Answer 1

首先 永远不要从远程同步加载数据 URL，始终使用异步方法，例如 URLSession.

'Any' has no subscript members

发生是因为编译器不知道中间对象是什么类型（例如 ["currently"]!["temperature"] 中的 currently）并且因为您使用的是 NSDictionary 编译器之类的基础集合类型完全不知道类型。

另外在Swift 3中需要通知编译器关于所有下标对象的类型。

您必须将 JSON 序列化的结果转换为实际类型。

此代码仅使用 URLSession 和 Swift 本机类型

let urlString = "https://api.forecast.io/forecast/apiKey/37.5673776,122.048951"

let url = URL(string: urlString)
URLSession.shared.dataTask(with:url!) { (data, response, error) in
  if error != nil {
    print(error)
  } else {
    do {

      let parsedData = try JSONSerialization.jsonObject(with: data!) as! [String:Any]
      let currentConditions = parsedData["currently"] as! [String:Any]

      print(currentConditions)

      let currentTemperatureF = currentConditions["temperature"] as! Double
      print(currentTemperatureF)
    } catch let error as NSError {
      print(error)
    }
  }

}.resume()

要打印 currentConditions 的所有键/值对，您可以编写

 let currentConditions = parsedData["currently"] as! [String:Any]

  for (key, value) in currentConditions {
    print("\(key) - \(value) ")
  }

关于jsonObject(with data的注释：

许多（似乎全部）教程建议 .mutableContainers 或 .mutableLeaves 选项，这在 Swift 中完全是胡说八道。这两个选项是遗留 Objective-C 选项，用于将结果分配给 NSMutable... 对象。在 Swift 中，任何 variable 默认情况下都是可变的，传递任何这些选项并将结果分配给 let 常量根本没有效果。此外，大多数实现永远不会改变反序列化的 JSON。

在 Swift 中唯一有用的（罕见）选项是 .allowFragments 如果 JSON 根对象可以是值类型（String , Number, Bool 或 null) 而不是其中一种集合类型 (array 或 dictionary)。但通常会省略 options 参数，这意味着 没有选项 。

=========================================== ==============================

解析的一些一般注意事项JSON

JSON是一种排列整齐的文本格式。读取 JSON 字符串非常容易。 仔细阅读字符串。只有六种不同的类型 – 两种集合类型和四种值类型。

集合类型是

数组-JSON：方括号中的对象[]-Swift：[Any]但在大多数情况下[[String:Any]]
字典 - JSON: 花括号中的对象 {} - Swift: [String:Any]

值类型是

String - JSON: 双引号中的任意值 "Foo", 甚至 "123"or "false" – Swift: String
数字 - JSON：数值不是双引号123或123.0 – Swift: Int 或 Double
Bool - JSON: true or false not in double quotes – Swift: true 或 false
null - JSON: null – Swift: NSNull

根据 JSON 规范，字典中的所有键都必须是 String。

基本上总是推荐使用可选绑定来安全地解包可选

如果根对象是字典 ({}) 将类型转换为 [String:Any]

if let parsedData = try JSONSerialization.jsonObject(with: data!) as? [String:Any] { ...

并通过键检索值（OneOfSupportedJSONTypes 是 JSON 集合或值类型，如上所述。）

if let foo = parsedData["foo"] as? OneOfSupportedJSONTypes {
    print(foo)
}

如果根对象是数组 ([]) 将类型转换为 [[String:Any]]

if let parsedData = try JSONSerialization.jsonObject(with: data!) as? [[String:Any]] { ...

并使用

遍历数组

for item in parsedData {
    print(item)
}

如果您需要特定索引处的项目，还要检查该索引是否存在

if let parsedData = try JSONSerialization.jsonObject(with: data!) as? [[String:Any]], parsedData.count > 2,
   let item = parsedData[2] as? OneOfSupportedJSONTypes {
      print(item)
    }
}

在极少数情况下，JSON 只是其中一种值类型，而不是集合类型，您必须传递 .allowFragments 选项并将结果转换为适当的值类型例如

if let parsedData = try JSONSerialization.jsonObject(with: data!, options: .allowFragments) as? String { ...

Apple 在 Swift 博客中发表了一篇综合文章：Working with JSON in Swift

=========================================== ==============================

在Swift 4+中，`Codable`协议提供了一种更方便的方法来将JSON直接解析为结构/类.

例如问题中给定的 JSON 样本（稍作修改）

let jsonString = """
{"icon": "partly-cloudy-night", "precipProbability": 0, "pressure": 1015.39, "humidity": 0.75, "precip_intensity": 0, "wind_speed": 6.04, "summary": "Partly Cloudy", "ozone": 321.13, "temperature": 49.45, "dew_point": 41.75, "apparent_temperature": 47, "wind_bearing": 332, "cloud_cover": 0.28, "time": 1480846460}
"""

可以解码成结构体Weather。 Swift 类型与上述相同。还有一些其他选项：

表示URL的字符串可以直接解码为URL。
time 整数可以用 dateDecodingStrategy .secondsSince1970.

Date

snaked_cased JSON 键可以转换为 camelCase 和 keyDecodingStrategy .convertFromSnakeCase

struct Weather: Decodable {
    let icon, summary: String
    let pressure: Double, humidity, windSpeed : Double
    let ozone, temperature, dewPoint, cloudCover: Double
    let precipProbability, precipIntensity, apparentTemperature, windBearing : Int
    let time: Date
}

let data = Data(jsonString.utf8)
do {
    let decoder = JSONDecoder()
    decoder.dateDecodingStrategy = .secondsSince1970
    decoder.keyDecodingStrategy = .convertFromSnakeCase
    let result = try decoder.decode(Weather.self, from: data)
    print(result)
} catch {
    print(error)
}

其他 Codable 来源：

Answer 2

Xcode 8 Beta 6 for Swift 3 发生的一个重大变化是 id 现在导入为 Any 而不是 AnyObject。

这意味着 parsedData 作为最有可能类型为 [Any:Any] 的字典返回。如果不使用调试器，我无法确切地告诉你你对 NSDictionary 的转换会做什么，但你看到的错误是因为 dict!["currently"]! 的类型为 Any

那么，你是如何解决这个问题的？从你引用它的方式来看，我假设 dict!["currently"]! 是一本字典，所以你有很多选择：

首先你可以这样做：

let currentConditionsDictionary: [String: AnyObject] = dict!["currently"]! as! [String: AnyObject]

这会给你一个字典对象，然后你可以查询值，这样你就可以像这样得到你的温度：

let currentTemperatureF = currentConditionsDictionary["temperature"] as! Double

或者，如果您愿意，可以排队：

let currentTemperatureF = (dict!["currently"]! as! [String: AnyObject])["temperature"]! as! Double

希望这有帮助，恐怕我还没有时间编写示例应用程序来测试它。

最后一点：最简单的方法可能是在开始时将 JSON 有效负载简单地转换为 [String: AnyObject]。

let parsedData = try JSONSerialization.jsonObject(with: data as Data, options: .allowFragments) as! Dictionary<String, AnyObject>

Answer 3

更新之后的isConnectToNetwork-Function，多亏了这个。

我为它写了一个额外的方法：

import SystemConfiguration

func loadingJSON(_ link:String, postString:String, completionHandler: @escaping (_ JSONObject: AnyObject) -> ()) {

    if(isConnectedToNetwork() == false){
        completionHandler("-1" as AnyObject)
        return
    }

    let request = NSMutableURLRequest(url: URL(string: link)!)
    request.httpMethod = "POST"
    request.httpBody = postString.data(using: String.Encoding.utf8)

    let task = URLSession.shared.dataTask(with: request as URLRequest) { data, response, error in
        guard error == nil && data != nil else { // check for fundamental networking error
            print("error=\(error)")
            return
        }

        if let httpStatus = response as? HTTPURLResponse , httpStatus.statusCode != 200 { // check for http errors
            print("statusCode should be 200, but is \(httpStatus.statusCode)")
            print("response = \(response)")
        }
        //JSON successfull
        do {
            let parseJSON = try JSONSerialization.jsonObject(with: data!, options: .allowFragments)
            DispatchQueue.main.async(execute: {
                completionHandler(parseJSON as AnyObject)
            });
        } catch let error as NSError {
            print("Failed to load: \(error.localizedDescription)")
        }
    }
    task.resume()
}

func isConnectedToNetwork() -> Bool {

    var zeroAddress = sockaddr_in(sin_len: 0, sin_family: 0, sin_port: 0, sin_addr: in_addr(s_addr: 0), sin_zero: (0, 0, 0, 0, 0, 0, 0, 0))
    zeroAddress.sin_len = UInt8(MemoryLayout.size(ofValue: zeroAddress))
    zeroAddress.sin_family = sa_family_t(AF_INET)

    let defaultRouteReachability = withUnsafePointer(to: &zeroAddress) {
        [=10=].withMemoryRebound(to: sockaddr.self, capacity: 1) {zeroSockAddress in
            SCNetworkReachabilityCreateWithAddress(nil, zeroSockAddress)
        }
    }

    var flags: SCNetworkReachabilityFlags = SCNetworkReachabilityFlags(rawValue: 0)
    if SCNetworkReachabilityGetFlags(defaultRouteReachability!, &flags) == false {
        return false
    }

    let isReachable = (flags.rawValue & UInt32(kSCNetworkFlagsReachable)) != 0
    let needsConnection = (flags.rawValue & UInt32(kSCNetworkFlagsConnectionRequired)) != 0
    let ret = (isReachable && !needsConnection)

    return ret
}

现在您可以在您的应用程序中随时随地轻松调用它

loadingJSON("yourDomain.com/login.php", postString:"email=\(userEmail!)&password=\(password!)") { parseJSON in

    if(String(describing: parseJSON) == "-1"){
        print("No Internet")
    } else {

    if let loginSuccessfull = parseJSON["loginSuccessfull"] as? Bool {
        //... do stuff
    }
}

Answer 4

let str = "{\"names\": [\"Bob\", \"Tim\", \"Tina\"]}"

let data = str.data(using: String.Encoding.utf8, allowLossyConversion: false)!

do {
    let json = try JSONSerialization.jsonObject(with: data, options: []) as! [String: AnyObject]
    if let names = json["names"] as? [String] 
{
        print(names)
}
} catch let error as NSError {
    print("Failed to load: \(error.localizedDescription)")
}

Answer 5

我正是为此而构建了 quicktype。只需粘贴您的示例 JSON，quicktype 就会为您的 API 数据生成此类型层次结构：

struct Forecast {
    let hourly: Hourly
    let daily: Daily
    let currently: Currently
    let flags: Flags
    let longitude: Double
    let latitude: Double
    let offset: Int
    let timezone: String
}

struct Hourly {
    let icon: String
    let data: [Currently]
    let summary: String
}

struct Daily {
    let icon: String
    let data: [Datum]
    let summary: String
}

struct Datum {
    let precipIntensityMax: Double
    let apparentTemperatureMinTime: Int
    let apparentTemperatureLowTime: Int
    let apparentTemperatureHighTime: Int
    let apparentTemperatureHigh: Double
    let apparentTemperatureLow: Double
    let apparentTemperatureMaxTime: Int
    let apparentTemperatureMax: Double
    let apparentTemperatureMin: Double
    let icon: String
    let dewPoint: Double
    let cloudCover: Double
    let humidity: Double
    let ozone: Double
    let moonPhase: Double
    let precipIntensity: Double
    let temperatureHigh: Double
    let pressure: Double
    let precipProbability: Double
    let precipIntensityMaxTime: Int
    let precipType: String?
    let sunriseTime: Int
    let summary: String
    let sunsetTime: Int
    let temperatureMax: Double
    let time: Int
    let temperatureLow: Double
    let temperatureHighTime: Int
    let temperatureLowTime: Int
    let temperatureMin: Double
    let temperatureMaxTime: Int
    let temperatureMinTime: Int
    let uvIndexTime: Int
    let windGust: Double
    let uvIndex: Int
    let windBearing: Int
    let windGustTime: Int
    let windSpeed: Double
}

struct Currently {
    let precipProbability: Double
    let humidity: Double
    let cloudCover: Double
    let apparentTemperature: Double
    let dewPoint: Double
    let ozone: Double
    let icon: String
    let precipIntensity: Double
    let temperature: Double
    let pressure: Double
    let precipType: String?
    let summary: String
    let uvIndex: Int
    let windGust: Double
    let time: Int
    let windBearing: Int
    let windSpeed: Double
}

struct Flags {
    let sources: [String]
    let isdStations: [String]
    let units: String
}

它还生成无依赖编组代码，以将 JSONSerialization.jsonObject 的 return 值哄骗到 Forecast，包括一个方便的构造函数，它接受 JSON 字符串，因此您可以快速解析强类型 Forecast 值并访问其字段：

let forecast = Forecast.from(json: jsonString)!
print(forecast.daily.data[0].windGustTime)

您可以使用 npm i -g quicktype 或 use the web UI 从 npm 安装 quicktype 以将完整的生成代码粘贴到您的 playground 中。

Answer 6

问题出在 API 交互方法上。 JSON 解析仅在语法上发生了变化。主要问题在于获取数据的方式。您使用的是同步获取数据的方式。这并非在所有情况下都有效。您应该使用的是一种异步方式来获取数据。这样就得通过API请求数据，然后等待它响应数据。您可以使用 URL 会话和 Alamofire 等第三方库来实现此目的。下面是 URL 会话方法的代码。

let urlString = "https://api.forecast.io/forecast/apiKey/37.5673776,122.048951"
let url = URL.init(string: urlString)
URLSession.shared.dataTask(with:url!) { (data, response, error) in
    guard error == nil else {
        print(error)
    }
    do {
        let Data = try JSONSerialization.jsonObject(with: data!) as! [String:Any]
        // Note if your data is coming in Array you should be using [Any]()
        //Now your data is parsed in Data variable and you can use it normally
        let currentConditions = Data["currently"] as! [String:Any]
        print(currentConditions)
        let currentTemperatureF = currentConditions["temperature"] as! Double
        print(currentTemperatureF)
    } catch let error as NSError {
        print(error)
    }
}.resume()

Answer 7

Swift 具有强大的类型推断。让我们摆脱“if let”或“guard let”样板文件并使用函数式方法强制解包：

这是我们的JSON。我们可以使用可选的 JSON 或通常的。我在我们的例子中使用可选的：

let json: Dictionary<String, Any>? = ["current": ["temperature": 10]]

辅助函数。我们只需要编写一次，然后在任何字典中重复使用：

/// Curry
public func curry<A, B, C>(_ f: @escaping (A, B) -> C) -> (A) -> (B) -> C {
    return { a in
        { f(a, [=11=]) }
    }
}

/// Function that takes key and optional dictionary and returns optional value
public func extract<Key, Value>(_ key: Key, _ json: Dictionary<Key, Any>?) -> Value? {
    return json.flatMap {
        cast([=11=][key])
    }
}

/// Function that takes key and return function that takes optional dictionary and returns optional value
public func extract<Key, Value>(_ key: Key) -> (Dictionary<Key, Any>?) -> Value? {
    return curry(extract)(key)
}

/// Precedence group for our operator
precedencegroup RightApplyPrecedence {
    associativity: right
    higherThan: AssignmentPrecedence
    lowerThan: TernaryPrecedence
}

/// Apply. g § f § a === g(f(a))
infix operator § : RightApplyPrecedence
public func §<A, B>(_ f: (A) -> B, _ a: A) -> B {
    return f(a)
}

/// Wrapper around operator "as".
public func cast<A, B>(_ a: A) -> B? {
    return a as? B
}

这是我们的魔法 - 提取值：

let temperature = (extract("temperature") § extract("current") § json) ?? NSNotFound

只需一行代码，无需强制解包或手动类型转换。此代码在 playground 中有效，因此您可以复制并检查它。这是一个实现 on GitHub.

Answer 8

这是解决您问题的另一种方法。因此，请查看以下解决方案。希望对你有帮助。

let str = "{\"names\": [\"Bob\", \"Tim\", \"Tina\"]}"
let data = str.data(using: String.Encoding.utf8, allowLossyConversion: false)!
do {
    let json = try JSONSerialization.jsonObject(with: data, options: []) as! [String: AnyObject]
    if let names = json["names"] as? [String] {
        print(names)
    }
} catch let error as NSError {
    print("Failed to load: \(error.localizedDescription)")
}

Answer 9

{
    "User":[
      {
        "FirstUser":{
        "name":"John"
        },
       "Information":"XY",
        "SecondUser":{
        "name":"Tom"
      }
     }
   ]
}

如果我使用以前的 json 创建模型使用此 link [博客]：http://www.jsoncafe.com 生成 Codable 结构或任何格式

型号

import Foundation
struct RootClass : Codable {
    let user : [Users]?
    enum CodingKeys: String, CodingKey {
        case user = "User"
    }

    init(from decoder: Decoder) throws {
        let values = try? decoder.container(keyedBy: CodingKeys.self)
        user = try? values?.decodeIfPresent([Users].self, forKey: .user)
    }
}

struct Users : Codable {
    let firstUser : FirstUser?
    let information : String?
    let secondUser : SecondUser?
    enum CodingKeys: String, CodingKey {
        case firstUser = "FirstUser"
        case information = "Information"
        case secondUser = "SecondUser"
    }
    init(from decoder: Decoder) throws {
        let values = try? decoder.container(keyedBy: CodingKeys.self)
        firstUser = try? FirstUser(from: decoder)
        information = try? values?.decodeIfPresent(String.self, forKey: .information)
        secondUser = try? SecondUser(from: decoder)
    }
}
struct SecondUser : Codable {
    let name : String?
    enum CodingKeys: String, CodingKey {
        case name = "name"
    }
    init(from decoder: Decoder) throws {
        let values = try? decoder.container(keyedBy: CodingKeys.self)
        name = try? values?.decodeIfPresent(String.self, forKey: .name)
    }
}
struct FirstUser : Codable {
    let name : String?
    enum CodingKeys: String, CodingKey {
        case name = "name"
    }
    init(from decoder: Decoder) throws {
        let values = try? decoder.container(keyedBy: CodingKeys.self)
        name = try? values?.decodeIfPresent(String.self, forKey: .name)
    }
}

解析

    do {
        let res = try JSONDecoder().decode(RootClass.self, from: data)
        print(res?.user?.first?.firstUser?.name ?? "Yours optional value")
    } catch {
        print(error)
    }

Answer 10

Swift 5 无法从您的 api 获取数据。解析 json 的最简单方法是使用 Decodable 协议。或者 Codable (Encodable & Decodable)。例如：

let json = """
{
    "dueDate": {
        "year": 2021,
        "month": 2,
        "day": 17
    }
}
"""

struct WrapperModel: Codable {
    var dueDate: DueDate
}

struct DueDate: Codable {
    var year: Int
    var month: Int
    var day: Int
}

let jsonData = Data(json.utf8)

let decoder = JSONDecoder()

do {
    let model = try decoder.decode(WrapperModel.self, from: jsonData)
    print(model)
} catch {
    print(error.localizedDescription)
}

正确解析 Swift 中的 JSON 3

Correctly Parsing JSON in Swift 3

parsing

json

swift

swift3

xcode8

解析的一些一般注意事项JSON

在Swift 4+中，`Codable`协议提供了一种更方便的方法来将JSON直接解析为结构/类.

正确解析 Swift 中的 JSON 3

Correctly Parsing JSON in Swift 3

parsing

json

swift

swift3

xcode8

解析的一些一般注意事项JSON

在Swift 4+中，Codable协议提供了一种更方便的方法来将JSON直接解析为结构/类.

在Swift 4+中，`Codable`协议提供了一种更方便的方法来将JSON直接解析为结构/类.