我真的在努力把一个JSON文件读入Swift,这样我就可以玩它了。我花了2天的时间重新搜索和尝试不同的方法,但没有运气,所以我已经注册了StackOverFlow,看看是否有人能给我指点正确的方向.....

我的JSON文件叫做test。Json,并包含以下内容:

{
  "person":[
     {
       "name": "Bob",
       "age": "16",
       "employed": "No"
     },
     {
       "name": "Vinny",
       "age": "56",
       "employed": "Yes"
     }
  ]
}    

该文件直接存储在文档中,我使用以下代码访问它:

let file = "test.json"
let dirs : String[] = NSSearchPathForDirectoriesInDomains(
                                                          NSSearchpathDirectory.DocumentDirectory,
                                                          NSSearchPathDomainMask.AllDomainMask,
                                                          true) as String[]

if (dirs != nil) {
    let directories: String[] = dirs
    let dir = directories[0]
    let path = dir.stringByAppendingPathComponent(file)
}

var jsonData = NSData(contentsOfFile:path, options: nil, error: nil)
println("jsonData \(jsonData)" // This prints what looks to be JSON encoded data.

var jsonDict = NSJSONSerialization.JSONObjectWithData(jsonData, options: nil, error: nil) as? NSDictionary

println("jsonDict \(jsonDict)") - This prints nil..... 

如果有人能给我一个正确的方向,我可以反序列化JSON文件,并把它放在一个可访问的Swift对象,我会永远感激!

亲切的问候,

Krivvenz。


当前回答

使用这个泛型函数

func readJSONFromFile<T: Decodable>(fileName: String, type: T.Type) -> T? {
    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
}

下面这行代码:

let model = readJSONFromFile(fileName: "Model", type: Model.self)

对于这种类型:

struct Model: Codable {
    let tall: Int
}

其他回答

Swiftyjson版本swift 3

func loadJson(fileName: String) -> JSON {

    var dataPath:JSON!

    if let path : String = Bundle.main.path(forResource: fileName, ofType: "json") {
        if let data = NSData(contentsOfFile: path) {
             dataPath = JSON(data: data as Data)
        }
    }
    return dataPath
}

Swift 5的答案为我工作,除了我必须添加一个空文件,重命名为xxx。Json,并使用泛型。

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

code

Swift 2.1答案(基于Abhishek):

    if let path = NSBundle.mainBundle().pathForResource("test", ofType: "json") {
        do {
            let jsonData = try NSData(contentsOfFile: path, options: NSDataReadingOptions.DataReadingMappedIfSafe)
            do {
                let jsonResult: NSDictionary = try NSJSONSerialization.JSONObjectWithData(jsonData, options: NSJSONReadingOptions.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 {}
    }

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。