Swift error: Reference to generic type Dictionary requires arguments in <...>

Question

错误 Reference to generic type Dictionary requires arguments in <...> 出现在函数的第一行。我试图让函数 return 从 api 中检索一个 NSDictionary。有人知道这里会发生什么吗？

class func getCurrentWeather(longitude: Float, latitude: Float)->Dictionary?{

let baseURL = NSURL(string: "https://api.forecast.io/forecast/\(apikey)/")
let forecastURL = NSURL(string: "\(longitude),\(latitude)", relativeToURL:baseURL)

let sharedSession = NSURLSession.sharedSession()
let downloadTask: NSURLSessionDownloadTask = sharedSession.downloadTaskWithURL(forecastURL!, completionHandler: { (location: NSURL!, response: NSURLResponse!, error: NSError!) -> Void in

    if(error == nil) {
        println(location)
        let dataObject = NSData(contentsOfURL:location!)
        let weatherDictionary: NSDictionary = NSJSONSerialization.JSONObjectWithData(dataObject!, options: nil, error: nil) as NSDictionary
        return weatherDictionary
    }else{
        println("error!")
        return nil

    }
})
}

编辑：

第二期：

    class func getCurrentWeather(longitude: Float, latitude: Float)->NSDictionary?{

    let baseURL = NSURL(string: "https://api.forecast.io/forecast/\(apikey)/")
    let forecastURL = NSURL(string: "\(longitude),\(latitude)", relativeToURL:baseURL)

    let sharedSession = NSURLSession.sharedSession()
    let downloadTask: NSURLSessionDownloadTask = sharedSession.downloadTaskWithURL(forecastURL!, completionHandler: { (location: NSURL!, response: NSURLResponse!, error: NSError!) -> Void in

        if(error == nil) {
            println(location)
            let dataObject = NSData(contentsOfURL:location!)
            let weatherDictionary: NSDictionary = NSJSONSerialization.JSONObjectWithData(dataObject!, options: nil, error: nil) as NSDictionary


            return weatherDictionary //ERROR: NSDictionary not convertible to void

        }else{
            println("error!")
            return nil ERROR: Type void does not conform to protocol 'NilLiteralConvertible'

        }
    })
    }

Answer 1

如果您打算 return 字典，那么您需要在其中指定键和数据的类型。

例如： 如果您的键和值都是字符串，那么您可以这样写：

class func getCurrentWeather(longitude: Float, latitude: Float) -> Dictionary <String, String>?
{
   ...
}

如果您不确定其中的数据或者您有多种类型的数据，请将 return 类型从 Dictionary 更改为 NSDictionary。

class func getCurrentWeather(longitude: Float, latitude: Float) -> NSDictionary?
{
    ...
}

或

你可以这样写：

class func getCurrentWeather(longitude: Float, latitude: Float) -> Dictionary <String, AnyObject>?
{
   ...
}

Swift error: Reference to generic type Dictionary requires arguments in <...>

Swift error: Reference to generic type Dictionary requires arguments in <...>

iphone

xcode

dictionary

ios

swift