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

Swift 4.1更新了Xcode 9.2

if let filePath = Bundle.main.path(forResource: "fileName", ofType: "json"), let data = NSData(contentsOfFile: filePath) {

     do {
      let json = try JSONSerialization.jsonObject(with: data as Data, options: JSONSerialization.ReadingOptions.allowFragments)        
        }
     catch {
                //Handle error
           }
 }

其他回答

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!
}

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
}

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)
        }
    }
}

我提供了另一个答案,因为这里没有一个答案是针对从测试包加载资源的。如果您正在使用一个输出JSON的远程服务,并且希望在不触及实际服务的情况下对解析结果进行单元测试,则可以获取一个或多个响应,并将它们放入项目中的Tests文件夹中的文件中。

func testCanReadTestJSONFile() {
    let path = NSBundle(forClass: ForecastIOAdapterTests.self).pathForResource("ForecastIOSample", ofType: "json")
    if let jsonData = NSData(contentsOfFile:path!) {
        let json = JSON(data: jsonData)
        if let currentTemperature = json["currently"]["temperature"].double {
            println("json: \(json)")
            XCTAssertGreaterThan(currentTemperature, 0)
        }
    }
}

这也使用了SwiftyJSON,但获得测试包和加载文件的核心逻辑是问题的答案。

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
}