无法分配给属性..可访问的斯威夫特 3 中的二传手

Question

我正在尝试从 Web API 获取艺术家信息并使用返回的数据创建一个新的艺术家类，但我无法将我的类属性设置为返回的数据。这是我的代码

进口基金会类曲目列表{

static var trackService:TrackService?
static func getArtistFromService()-> Artist
{
    // construct the complete URL endpoint ot be called
    //let searchURL = "https://(siteURL)/?s=(searchTerm)"
    var searchedartist : Artist
    let searchURL = "http://ws.audioscrobbler.com/2.0/?method=artist.gettoptracks&artist=drake&api_key=d932ba18a79b1739a4d9ba02c63bdb61&format=json"
    // degugging - can test in browser if call is failing
    print ("Web Service call = (searchURL)")
    // create the Web Service object that will make the call - initialiser
    trackService = TrackService(searchURL)
    // create an operation queue - to allow the web service to be called on a secondary thread...
    let operationQ = OperationQueue()
    //... will only allow one task at a time in queue...
    operationQ.maxConcurrentOperationCount = 1
    //...queue that web service object as the operation object to be run on this thread
    operationQ.addOperation(trackService!)
    //... wait for the operation to complete [Synchronous] - better to use delegation, for [Asynchronous]
    operationQ.waitUntilAllOperationsAreFinished()
    // clear the current list of movies
    // get the raw JSON back from the completed service call [complete dataset]
    let returnedJSON = trackService!.jsonFromResponse
    let artist = returnedJSON?["artist"] as! String
    let bio = returnedJSON?["bio"] as! String
          searchedartist.name = artist
        searchedartist.bio = bio

    // return the main list of Movie objects to caller (UI)
    return searchedartist
}

// for debugging - displays some details of the Movies currently in the main list...

}//类声明结束

类艺术家{ 私有(设置(变量名称：字符串 私有 (设置( 变量 生物：字符串

init?(_ n:String, _ b:String)
{
    name = n
    bio = b
}
convenience init(_ JSONObject:[String:String])
{
    let name = JSONObject["name"]!
    let bio = JSONObject["bio"]!
    self.init(name, bio)!
}

}

Answer 1

您的类有私有二传手。只需使用返回的数据创建一个新艺术家：

let artist = returnedJSON.....
let bio = returnedJSON.....
let searchedArtist = Artist(artist, bio)
return searchedArtist

或者只是使用您的便利初始化。一般来说，我更喜欢拥有不可变的模型对象，所以我不建议公开你的 setter 的替代方案。