Alamofire 4从“(_) throws ->()”类型的抛出函数到非抛出函数类型“(DataResponse <Any>)-> Void”的无效转换
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【中文标题】Alamofire 4从“(_) throws ->()”类型的抛出函数到非抛出函数类型“(DataResponse <Any>)-> Void”的无效转换【英文标题】:Alamofire 4 Invalid conversion from throwing function of type '(_) throws -> ()' to non-throwing function type '(DataResponse<Any>) -> Void' 【发布时间】:2018-07-31 15:01:49 【问题描述】:我刚刚运行了 pod update 命令。并在我的 Alamofire 请求的 .responseJSON response in
块中收到此错误。
'(_) throws -> ()' 类型的抛出函数的无效转换 到非抛出函数类型 '(DataResponse) -> Void'
这是截图
更新 1
这是我的代码
Alamofire.request(getPublicKeyUrl!, method: .get, parameters: nil, encoding: JSONEncoding.default)
.downloadProgress(queue: DispatchQueue.global(qos: .utility)) progress in
print("Progress: \(progress.fractionCompleted)")
.validate request, response, data in
// Custom evaluation closure now includes data (allows you to parse data to dig out error messages if necessary)
//print("response", response);
return .success
.responseJSON response in
//debugPrint(response)
//print("JSON req", response)
if((response.result.value) != nil)
let swiftyJSON = JSON(response.result.value!)
print(swiftyJSON)
let Code = swiftyJSON["LL"]["Code"].stringValue
print("response Code", Code);
if (Code == "200")
// get key
self._publicKey = swiftyJSON["LL"]["value"].stringValue
print("publicKey", self._publicKey);
var request = URLRequest(url: (URL(string:"ws://\(self._ip)/ws/rfc6455"))!)
print("Request", String(describing: request.url))
request.setValue("websocket", forHTTPHeaderField: "Upgrade")
request.setValue("Upgrade", forHTTPHeaderField: "Connection")
request.setValue("remotecontrol", forHTTPHeaderField: "Sec-WebSocket-Protocol")
request.setValue(self._hash as String, forHTTPHeaderField: "Sec-WebSocket-Key")
self.socket = WebSocket(request: request)
// self.commandInPending.insert("authenticate/\(self._hash)", at: 0)
// self.socket.delegate = self
// self.socket.connect()
// check AES encryption
//let message = "Don´t try to read this text. Top Secret Stuff"
//let messageData = message.data(using:String.Encoding.utf8)!
let keyData = "12345678901234567890123456789012".data(using:String.Encoding.utf8)!
let ivData = "abcdefghijklmnop".data(using:String.Encoding.utf8)!
print("keyData", keyData);
print("ivData", ivData);
let key = keyData.map String(format:"%02x", $0) .joined()
let iv = ivData.map String(format:"%02x", $0) .joined()
print("keyHex", key);
print("ivData", iv);
// let session_key =
let keyAndiv = ("\(key):\(iv)") // self._publicKey
print("keyAndiv", keyAndiv);
// let str = "Clear Text"
let clear = try ClearMessage(string: keyAndiv, using: .utf8)
let encrypted = try clear.encrypted(with: publicKey, padding: .PKCS1)
let data = encrypted.data
let base64String = encrypted.base64Encoded
print ("data", data);
print ("base64String", base64String);
//let encryptedData = AppUtils.testCrypt(data:messageData, keyData:keyData, ivData:ivData, operation:kCCEncrypt)
//let decryptedData = AppUtils.testCrypt(data:encryptedData, keyData:keyData, ivData:ivData, operation:kCCDecrypt)
//let decrypted = String(bytes:decryptedData, encoding:String.Encoding.utf8)!
//print("message", message);
//print("decrypted", decrypted);
有什么线索吗?
【问题讨论】:
请贴代码而不是图片,这样我们可以清楚地理解。但问题是因为您在responseJSON
完成处理程序中调用了一个抛出函数。查看WebSocket Library 和SwiftyJSON,看看你是否使用过任何投掷函数。没有您的代码,我无法提供进一步的帮助
@PranavKasetti 抱歉耽搁了。上面添加的代码。谢谢
【参考方案1】:
发生错误是因为您没有处理throw
ing 函数的错误。
添加do - catch block
do
let clear = try ClearMessage(string: keyAndiv, using: .utf8)
let encrypted = try clear.encrypted(with: publicKey, padding: .PKCS1)
let data = encrypted.data
let base64String = encrypted.base64Encoded
print ("data", data);
print ("base64String", base64String)
catch print(error)
这是 Swift,if
条件和尾随分号不需要括号
if response.result.value != nil ...
或更好
guard let result = response.result.value else return
let swiftyJSON = JSON(result)
【讨论】:
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