Swift - 使用alamofire发布请求

问题描述 投票:0回答:1

我对Alamofire有点麻烦。我有成功在Android设备上工作的服务器功能。一切正常,所有数据都可以正常运行。但是每当我发送带有2个以上参数的post请求时,我都会在服务器上收到错误,因此无法发布数据。我试图将post作为字符串放入自定义编码中,因为当我将参数放入Alamofire.request方法时,我也遇到了错误,但服务器响应是“错误处理程序中的错误”。由于String扩展,我将所有JSON更改为字符串,然后我将每个“[”和“]”更改为“{”,“}”,但仍然无法将数据保存到服务器。

编码字典并在Swift中更改为JSON,如下所示:

%22%22%22%22%22%22%22%22%22%22%22%22%22%22%22%22%3%22% 22%22%22%3%22%22%22%22%22%22%22%22%

Swift中的字典

["calls": ["0": ["call_date": "2018-06-27+11:49:18", "number": "56262621", "sms": "0", "duration": "0", "type": 211]]]

另一方面,Android上的相同应用程序可以完美地使用相同的数据。

Java中的编码JSON如下所示:

%22%22%22%22%22%22%22%22%22%22%22%22%22%22%22%22%22%22% %22%22%22%22%22%22%22%22%22%22%

Java中的JSON

{"calls":{"0":{"number":"852486258965","duration":"0","type":211,"call_date":"2018-06-27 11:53:53","sms":"0"}}}

为什么在Android应用程序中一切正常但在iOS上它根本不起作用?有没有人有这个想法?

Swift中的函数

func sendStoreCalls(parameters: Dictionary<Int, Dictionary<String, Any>>) {

        var params: Parameters = [:]
        var par: Parameters = [:]
        var pa: [String:Any] = [:]
        var i = 0
        for (_, param) in parameters {
            if param["isSend"] as! Int == 0 {
                par["duration"] = param["callTime"]
                par["sms"] = "0"
                let callDate: String = param["callDate"] as! String
                par["type"] = param["callType"]
                par["call_date"] = callDate.replacingOccurrences(of: " ", with: "+", options: .literal, range: nil)
                if param["callNumber"] as! String != "" {
                    par["number"] = param["callNumber"]
                }
                params[String(i)] = par
                i += 1
            }
        }

        pa["calls"] = params

        let token: String = SharedPreferences.getPrefToken()!

        let bufferedParams = pa.queryParameters

        let transformedPar = bufferedParams.replacingOccurrences(of: "%5B", with: "%7B", options: .literal, range: nil).replacingOccurrences(of: "%5D", with: "%7D", options: .literal, range: nil)

        let trn = transformedPar.replacingOccurrences(of: "%C2A0", with: "", options: .literal, range: nil).replacingOccurrences(of: "%20", with: "", options: .literal, range: nil)

        let url = "https://url.com?api_token=\(token)"

        Alamofire.request(url, method: .post, encoding: JSONStringArrayEncoding.init(string: trn))
            .responseJSON {
                response in
                print(NSString(data: (response.request?.httpBody)!, encoding: String.Encoding.utf8.rawValue)!)
                switch response.result {
                case .success(let value):
                    let json = JSON(value)
                    print(json["message"].rawString()!)
                case .failure(let error):
                    if let data = response.data {
                        print("ERROR!!!")
                        print(String(data: data, encoding: String.Encoding.utf8)!)
                    }
                    print(error)
                }
            }
    }
ios json swift
1个回答
2
投票

尝试使用/不使用编码参数的方法以及添加/删除标题[“Content-Type”:“application / json”]

func request(_ method: HTTPMethod
    , _ URLString: String
    , parameters: [String : AnyObject]? = [:]
    , headers: [String : String]? = [:]
    , onView: UIView?, vc: UIViewController, completion:@escaping (Any?) -> Void
    , failure: @escaping (Error?) -> Void) {


    Alamofire.request(URLString, method: method, parameters: parameters, encoding: JSONEncoding.default, headers: headers)
        .responseJSON { response in

            switch response.result {
            case .success:
                completion(response.result.value!)
            case .failure(let error):
                failure(error)
                guard error.localizedDescription == JSON_COULDNOT_SERIALISED else {
                    AppUtil().showMessage((error.localizedDescription), messageTitle: EMPTY_STRING, buttonTitle: OK, vc: vc)
                    return
                }
                //                    AppUtil.showMessage(SOMETHING_WNET_WRONG, messageTitle: EMPTY_STRING, buttonTitle: OK)
            }
    }
}
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