这在XCode 8.3.3中是可行的

func fetchPersons(){

    if let pathURL = Bundle.main.url(forResource: "Person", withExtension: "json"){

        do {

            let jsonData = try Data(contentsOf: pathURL, options: .mappedIfSafe)

            let jsonResult = try JSONSerialization.jsonObject(with: jsonData, options: .mutableContainers) as! [String: Any]
            if let persons = jsonResult["person"] as? [Any]{

                print(persons)
            }

        }catch(let error){
            print (error.localizedDescription)
        }
    }
}

在清理和抛光我的代码之后，我来到了这两个函数，你可以添加到你的项目中，并使用它们非常整洁和快速地从json文件读取数据，并将数据转换为你想要的任何类型!

public func readDataRepresentationFromFile(resource: String, type: String) -> Data? {
    let filePath = Bundle.main.path(forResource: resource, ofType: type)
    
    if let path = filePath {
        let result = FileManager.default.contents(atPath: path)
        return result
    }
    return nil
}

然后在这个函数的帮助下，你可以将你的数据转换为任何你想要的类型:

public func getObject<T: Codable>(of type: T.Type, from file: String) -> T?  {
    guard let data = readDataRepresentationFromFile(resource: file, type: "json") else {
        return nil
    }
    if let object = try? JSONDecoder().decode(type, from: data) {
        return object
    }
    return nil
}

此代码的应用示例: 在你的代码中调用这个函数，给它你的json文件的名字，这就是你所需要的!

func getInputDataFromSomeJson(jsonFileName: String) -> YourReqiuredOutputType? {
    return getObject(of: YourReqiuredOutputType.self, from: jsonFileName)
}

Swift 4 JSON类与可解码-为那些喜欢类

定义类如下:

class People: Decodable {
  var person: [Person]?

  init(fileName : String){
    // url, data and jsonData should not be nil
    guard let url = Bundle.main.url(forResource: fileName, withExtension: "json") else { return }
    guard let data = try? Data(contentsOf: url) else { return }
    guard let jsonData = try? JSONDecoder().decode(People.self, from: data) else { return }

    // assigns the value to [person]
    person = jsonData.person
  }
}

class Person : Decodable {
  var name: String
  var age: String
  var employed: String
}

用法，非常抽象:

let people = People(fileName: "people")
let personArray = people.person

这允许People类和Person类的方法，如果需要，变量(属性)和方法也可以标记为private。

为Swift 3更新了最安全的方式

    private func readLocalJsonFile() {

    if let urlPath = Bundle.main.url(forResource: "test", withExtension: "json") {

        do {
            let jsonData = try Data(contentsOf: urlPath, options: .mappedIfSafe)

            if let jsonDict = try JSONSerialization.jsonObject(with: jsonData, options: .mutableContainers) as? [String: AnyObject] {

                if let personArray = jsonDict["person"] as? [[String: AnyObject]] {

                    for personDict in personArray {

                        for (key, value) in personDict {

                            print(key, value)
                        }
                        print("\n")
                    }
                }
            }
        }

        catch let jsonError {
            print(jsonError)
        }
    }
}

//change type based on your struct and right JSON file

let quoteData: [DataType] =
    load("file.json")

func load<T: Decodable>(_ filename: String, as type: T.Type = T.self) -> T {
    let data: Data

    guard let file = Bundle.main.url(forResource: filename, withExtension: nil)
        else {
            fatalError("Couldn't find \(filename) in main bundle.")
    }

    do {
        data = try Data(contentsOf: file)
    } catch {
        fatalError("Couldn't load \(filename) from main bundle:\n\(error)")
    }

    do {
        let decoder = JSONDecoder()
        return try decoder.decode(T.self, from: data)
    } catch {
        fatalError("Couldn't parse \(filename) as \(T.self):\n\(error)")
    }
}

根据Abhishek的回答，对于iOS 8，这将是:

let masterDataUrl: NSURL = NSBundle.mainBundle().URLForResource("masterdata", withExtension: "json")!
let jsonData: NSData = NSData(contentsOfURL: masterDataUrl)!
let jsonResult: NSDictionary = NSJSONSerialization.JSONObjectWithData(jsonData, options: nil, error: nil) as! NSDictionary
var persons : NSArray = jsonResult["person"] as! NSArray