Question

我正在尝试检查json响应的字典之一的空条件。但是当我检查条件时，出现类似的错误“二进制运算符'=='不能应用于类型的操作数'[[String：Any]'和'JSON'“ 。我在if !(chattt == JSON.null)行遇到此错误。如果有人能帮助我解决此问题，那就太好了。谢谢！ / p>

 let acce:String = UserDefaults.standard.string(forKey: "access-tokenn")!
        print(acce)

        let headers:HTTPHeaders = ["Authorization":"Bearer \(acce)","Content-Type":"application/X-Access-Token"]
        print((Constants.Chatlistsearch)+(idd))
        Alamofire.request((Constants.Chatlistsearch+idd), method: .get, encoding: URLEncoding.default, headers: headers).responseJSON {  response in
            switch response.result {
            case .success:
                //print(response)
                if response.result.value != nil{
                    var maindictionary = NSDictionary()
                    maindictionary = response.result.value as! NSDictionary
                  //  print(maindictionary)

                    var chat:Dictionary = maindictionary.value(forKey: "data") as! [String:Any]
                   // print(chat)

                    var chatt:Dictionary = chat["user"] as! [String:Any]
                   // print(chatt)

                    var chattt:Dictionary = chat["chat"] as! [String:Any]
                    print(chattt)



                    if !(chattt == JSON.null) {
                        let viewc = UIStoryboard.init(name: "Main", bundle: Bundle.main).instantiateViewController(withIdentifier: "ChatViewController") as? ChatViewController
                        self.navigationController?.pushViewController(viewc!, animated: true)
                    }else{
                        print("Find Action")
                    }


//                    self.data = [String(stringInterpolationSegment: chatt["unique_id"])]
//                    print(self.data)

                }
                break
            case .failure(let error):

                print(error)
            }
        }

    }

Answer 1

您可以这样做

if let chattt = chat["chat"] as? [String:Any] {
    //It's available. Execute further tasks
} else {
    //It's nil or chat["chat"] type is not dictionary.
}

您可以使用后卫来做。

guard let chattt = chat["chat"] as? [String:Any] else {
    //Using this, the further code will never execute as implemented force return here
    return
}

如果有任何问题，请告诉我。

Answer 2

执行以下操作

guard let chattt = chat["chat"] as? [String:Any] else {
    let viewc = UIStoryboard.init(name: "Main", bundle: Bundle.main).instantiateViewController(withIdentifier: "ChatViewController") as? ChatViewController
    self.navigationController?.pushViewController(viewc!, animated: true)
    return
}

如何检查JSON响应的空条件？

2 个答案: