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I want to send more than one "davetiyeID". For this, I send the DAVMED_ID array as a parameter. But when printing, I get blank output. What do I need to do to send more than one value? So I want to send more than one davetiyeID. Is it possible for me to do this? Can I send 3-4 invitation IDs instead of one davetiyeID?

extension Array {
    /// Convert the receiver array to a `Parameters` object.
    func asParameters() -> Parameters {
        return [arrayParametersKey: self]
    }
}

var DAVMED_ID : [String] = []

  public struct ArrayEncoding: ParameterEncoding {

        public let options: JSONSerialization.WritingOptions

        public init(options: JSONSerialization.WritingOptions = []) {
            self.options = options
        }

        public func encode(_ urlRequest: URLRequestConvertible, with parameters: Parameters?) throws -> URLRequest {
            var urlRequest = try urlRequest.asURLRequest()

            guard let parameters = parameters,
                let array = parameters[arrayParametersKey] else {
                    return urlRequest
            }

            do {
                let data = try JSONSerialization.data(withJSONObject: array, options: options)

                if urlRequest.value(forHTTPHeaderField: "Content-Type") == nil {
                    urlRequest.setValue("application/json", forHTTPHeaderField: "Content-Type")
                }

                urlRequest.httpBody = data

            } catch {
                throw AFError.parameterEncodingFailed(reason: .jsonEncodingFailed(error: error))
            }

            return urlRequest
        }
    }

 @objc func favKontrol(){
  let url = URL(string: "...")!
  var request = URLRequest(url: url)

 request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
 request.httpMethod = "POST"
  Alamofire.request(
        url,
        method: .post,
        parameters: DAVMED_ID.asParameters(),
        encoding: ArrayEncoding(),
        headers: nil).responseJSON { (responseData) -> Void in

            switch (responseData.result) {
            case .success(let result):
...
}
deneov
  • 63
  • 6

0 Answers0