ios - 无法在 swift 3 中获得正确的 api 调用

Question

我在 postman 中尝试了这个特定的 API，它返回了一个很好的结果。

因此，我创建了一个参数

["JsonRequest": "{\"header\":\"GetLocationListReq\",\"accessKey\":\"fakeKey\"}"]

但是当我在我的 Xcode 项目中使用 swift 和 alamofire 调用这个 API 时传递这个参数时，我总是得到一个错误。

有兴趣的 friend ，我的apiRouter是

    //
//  WINAPIRouter.swift
//  Winner21
//
//  Created by Lin Hairui on 27/4/17.
//  Copyright © 2017 Pioneers & Leaders (Publishers). All rights reserved.
//

import Foundation
import Alamofire

enum WINAPIRouter : URLRequestConvertible {
    static let baseURL = Constant.baseURL

    case get(String)
    case create([String:Any],String?)
    case delete(String)

    func asURLRequest() throws -> URLRequest {

        var method : HTTPMethod {
            switch self {
            case .get:
                return HTTPMethod.get
            case .create:
                return HTTPMethod.post
            case .delete:
                return HTTPMethod.delete
            }
        }


        let params:(Dictionary<String, Any>?) = {
            switch self {
            case .get, .delete:
                return nil
            case .create(let params, _):
                var fieldParams = params
                fieldParams["accessKey"] = Constant.kAPIAccessKey

                let jsonData: NSData
                do {
                    jsonData = try JSONSerialization.data(withJSONObject: fieldParams, options: JSONSerialization.WritingOptions()) as NSData
                    let jsonString = NSString(data: jsonData as Data, encoding: String.Encoding.utf8.rawValue) as! String
                    let authParam : [String : Any] = [
                        "JsonRequest" : jsonString
                    ]
                    print("google was here \(authParam)")
                    return authParam

                } catch _ {
                    print ("JSON Failure")
                }
                return nil
            }
        }()

        let url : URL = {
            return URL(string: Constant.baseURL)!
        }()

        var urlRequest = URLRequest(url: url)


//        urlRequest.addValue("application/json", forHTTPHeaderField: "Content-Type")
//        urlRequest.addValue("application/json", forHTTPHeaderField: "Accept")
        urlRequest.httpMethod = method.rawValue
        print("facebook was here = \(urlRequest.httpMethod!)")

        let encoding = JSONEncoding.default
        print("yahoo was here \(params)")
        //let facejsonData = try! JSONSerialization.data(withJSONObject: params!, options: JSONSerialization.WritingOptions())
        return try encoding.encode(urlRequest, with: params!)
    }

}

并且，我使用以下函数调用 API。

Alamofire.request(WINAPIRouter.create(params, nil)).responseJSON

Answer 1

我认为来自 API 的响应不是 JSON 格式，请尝试使用如下 URL 编码，

Alamofire.request(path, method: .get, parameters: parameters, encoding: URLEncoding.default, headers: nil) . responseString(completionHandler: { ( dataResponse ) in
                 /// print response

            })