这对我很有效

func readjson(fileName: String) -> NSData{

    let path = NSBundle.mainBundle().pathForResource(fileName, ofType: "json")
    let jsonData = NSData(contentsOfMappedFile: path!)

    return jsonData!
}

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。

我使用下面的代码从FAQ-data中获取JSON。Json文件存在于项目目录中。

我在Xcode 7.3中使用Swift实现。

     func fetchJSONContent() {
            if let path = NSBundle.mainBundle().pathForResource("FAQ-data", ofType: "json") {

                if let jsonData = NSData(contentsOfFile: path) {
                    do {
                        if let jsonResult: NSDictionary = try NSJSONSerialization.JSONObjectWithData(jsonData, options: NSJSONReadingOptions.MutableContainers) as? NSDictionary {

                            if let responseParameter : NSDictionary = jsonResult["responseParameter"] as? NSDictionary {

                                if let response : NSArray = responseParameter["FAQ"] as? NSArray {
                                    responseFAQ = response
                                    print("response FAQ : \(response)")
                                }
                            }
                        }
                    }
                    catch { print("Error while parsing: \(error)") }
                }
            }
        }

override func viewWillAppear(animated: Bool) {
        fetchFAQContent()
    }

JSON文件结构:

{
    "status": "00",
    "msg": "FAQ List ",
    "responseParameter": {
        "FAQ": [
            {                
                "question": “Question No.1 here”,
                "answer": “Answer goes here”,  
                "id": 1
            },
            {                
                "question": “Question No.2 here”,
                "answer": “Answer goes here”,
                "id": 2
            }
            . . .
        ]
    }
}

简化Peter Kreinz提供的例子。适用于Swift 4.2。

扩展函数:

extension Decodable {
  static func parse(jsonFile: String) -> Self? {
    guard let url = Bundle.main.url(forResource: jsonFile, withExtension: "json"),
          let data = try? Data(contentsOf: url),
          let output = try? JSONDecoder().decode(self, from: data)
        else {
      return nil
    }

    return output
  }
}

示例模型:

struct Service: Decodable {
  let name: String
}

示例用法:

/// service.json
/// { "name": "Home & Garden" }

guard let output = Service.parse(jsonFile: "service") else {
// do something if parsing failed
 return
}

// use output if all good

这个例子也适用于数组:

/// services.json
/// [ { "name": "Home & Garden" } ]

guard let output = [Service].parse(jsonFile: "services") else {
// do something if parsing failed
 return
}

// use output if all good

注意，我们没有提供任何不必要的泛型，因此不需要强制转换parse的结果。

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)
}

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