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)
        }
    }
}

Xcode 8 Swift 3读取json文件更新:

    if let path = Bundle.main.path(forResource: "userDatabseFakeData", ofType: "json") {
        do {
            let jsonData = try NSData(contentsOfFile: path, options: NSData.ReadingOptions.mappedIfSafe)
            do {
                let jsonResult: NSDictionary = try JSONSerialization.jsonObject(with: jsonData as Data, options: JSONSerialization.ReadingOptions.mutableContainers) as! NSDictionary
                if let people : [NSDictionary] = jsonResult["person"] as? [NSDictionary] {
                    for person: NSDictionary in people {
                        for (name,value) in person {
                            print("\(name) , \(value)")
                        }
                    }
                }
            } catch {}
        } catch {}
    }

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)
}

我浪费了我的时间在定位文件，它位于我的项目与名称Jsondata.json。但是我无法通过代码....找到我的文件

解决方案:确保您的Jsondata。在项目>构建阶段>复制Bundle资源中添加了json文件。否则你将无法获得file和Bundle.main。url(forResource: fileName, witheextension: "json")将总是给你nil。

一般的方法可以是这样的:

创建响应类名称字符串的json文件

struct Response: Codable,FileDecodable {
    typealias T = Self
    let names:[Data]
}
protocol FileDecodable{
   associatedtype T:Codable
   static func loadJson() ->T?
}

extension FileDecodable{
    static func loadJson() -> T? {
        let fileName = String(describing: T.self)
        if let url = Bundle.main.url(forResource: fileName, withExtension: "json")     {
            do {
                let data = try Data(contentsOf: url)
                let decoder = JSONDecoder()
                let jsonData = try decoder.decode(T.self, from: data)
                return jsonData
            } catch {
                print("error:\(error)")
            }
        }
        return nil
    }
}

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。