如何获取函数结果并将其转换为键值对数组

时间:2020-02-26 16:07:14

标签: arrays swift dictionary

我对Swift还是很陌生,但是我有一个函数可以返回数字的键值对结果

func dataPostRequest(_ url:String,_ paramString:String)
{
    let url:NSURL = NSURL(string: url)!
    let session = URLSession.shared

    let request = NSMutableURLRequest(url: url as URL)
    request.httpMethod = "POST"

    request.httpBody = paramString.data(using: String.Encoding.utf8)

    let task = session.dataTask(with: request as URLRequest) {
        (
        data, response, error) in

        guard let _:NSData = data as NSData?, let _:URLResponse = response, error == nil else {
            print("error")
            return
        }

        if let dataString = NSString(data: data!, encoding: String.Encoding.utf8.rawValue)
        {
            print(dataString)


        }
    }

    task.resume()

}

我可以这样调用函数:

dataPostRequest("https://myserver.com/","user_email=emailtest.com")

这可以正常工作,但是我现在想使用提取的数据并将其显示在“表视图”中。我的tableView设置正确,但是我想知道如何使用函数并将其转换为可以使用的键值对数组或字典。我尝试创建一个空字典并将其函数调用设置为它,但是出现错误:

var emptyDictionary = [Int: Int]()

emptyDictionary = dataPostRequest("https://myserver.com/","user_email=emailtest.com")

无论我尝试什么变化,我都会不断得到:

Cannot assign value of type '()' to type '[Int : Int]' 

任何指导将不胜感激。

1 个答案:

答案 0 :(得分:1)

dataPostRequest没有返回值(aka ())。您可以在完成处理程序中解码接收到的数据,并将其分配给字典。请参阅下面的评论行。

如果需要继续执行其他功能,则必须添加描述为here的完成处理程序。

如果有本机对应,则基本上根本不会在Swift中使用NS...类。并且不要注释编译器可以推断的类型。

Swift的原生版本是

func dataPostRequest(with url:String, paramString : String)
{
    let url = URL(string: url)!
    let session = URLSession.shared

    var request = URLRequest(url: url)
    request.httpMethod = "POST"

    request.httpBody = paramString.data(using: .utf8)

    let task = session.dataTask(with: request) { data, response, error in

        if let error = error {
            print(error)
            return
        }

        let dataString = String(data: data!, encoding: .utf8)!
        print(dataString)

        // here decode the data to the desired type and assign it to emptyDictionary
    }

    task.resume()
}