Json 解析错误：在 Swift iOS 中使用未声明的类型“Foundation”

Question

我正在使用 Swift 创建一个应用程序，我必须使用以下 URL 显示国家列表：

https://restcountries.eu/rest/v2/all

在这里我只需要获取name 密钥并添加到我的NSMutableArray。当我尝试使用 URLSession 解析它时，出现错误：

expression produced error: error: /var/folders/_3/27lhgzw9699c4_yg37r72_d80000gp/T/expr40-bcf47d..swift:1:65: error: use of undeclared type 'Foundation'
Swift._DebuggerSupport.stringForPrintObject(Swift.UnsafePointer<Foundation.Data>(bitPattern: 0x105c0c2f0)!.pointee)

下面是我的代码：

func fetchCountryList(countryURL:URL, completion:@escaping (NSDictionary) -> ()) {

            print(countryURL)

            let request = NSMutableURLRequest( url: countryURL as URL)

            let task = URLSession.shared.dataTask(with: request as URLRequest) {
                data, response, error in
                do{
                    if let data = data,
                        let jsonString =  try JSONSerialization.jsonObject(with: data, options: JSONSerialization.ReadingOptions.mutableContainers) as? NSDictionary
                        , error == nil {
                        completion(jsonString)
                    } else {
                        print("error=\(error!.localizedDescription)")
                        let errorDict = ["error_status":true,"message":error!.localizedDescription] as [String : Any]
                        completion(errorDict as NSDictionary)

                    }
                }
                catch{
                    print("error=\(error.localizedDescription)")
                    let errorDict = ["error_status":true,"message":error.localizedDescription] as [String : Any]
                    completion(errorDict as NSDictionary)

                }

            }
            task.resume()
    }

任何人都可以在这里建议我我的错误。提前致谢！

Answer 1

替换

try JSONSerialization.jsonObject(with: data, options: JSONSerialization.ReadingOptions.mutableContainers) as? NSDictionary

有

try JSONSerialization.jsonObject(with: data, options: JSONSerialization.ReadingOptions.mutableContainers) as? [[String: Any]]

由于它是一个对象数组，我们还需要将 json 转换为 array

要获得array of country names，我们可以这样做：

guard let countries = try JSONSerialization.jsonObject(with: data, options: JSONSerialization.ReadingOptions.mutableContainers) as? [[String: Any]] else {
    return
}
let countryNames = countries.map { $0["name"] as! String }

Answer 2

替换

 let jsonString =  try JSONSerialization.jsonObject(with: data, options: JSONSerialization.ReadingOptions.mutableContainers) as? NSDictionary
                

与

if let result =  try JSONSerialization.jsonObject(with: data, options:[]) as? [[String:Any]] {
   for item in result {
     print(item["name"])
   }  
 }

响应是数组而不是字典也不要使用 NS 的东西

NSDictionary => [字符串：任意]

NSUrl => 网址

NSMutableURLRequest => URLRequest

---

func fetchCountryList(countryURL:URL, completion:@escaping (_ arr :[[String:Any]]?, _ error:Error?) -> ()) {
        
            print(countryURL)
        
            let request = URLRequest( url: countryURL)

            let task = URLSession.shared.dataTask(with: request) {
                data, response, error in
                do{
                    if let data = data,
                        let jsonString =  try JSONSerialization.jsonObject(with: data, options:[]) as? [[String:Any]]
                        , error == nil {
                        completion(jsonString,nil)
                    } else {
                        print("error=\(error!.localizedDescription)")
                        let errorDict = ["error_status":true,"message":error!.localizedDescription] as [String : Any]
                        completion(nil,error)
                        
                    }
                }
                catch{
                    print("error=\(error.localizedDescription)")
                    let errorDict = ["error_status":true,"message":error.localizedDescription] as [String : Any]
                    completion(nil,error)
                    
                }
                
            }
            task.resume()
    }

---

打电话

serviceModel.fetchCountryList(countryURL:countryURL){ (result,error) in 
  if let res = result {
    print(res)
  } 
}