采用 Swift HTTP POST 并解码 JSON 响应

Question

根据之前的一些帮助，我能够构建它以从 swift 平台内提交 API POST 请求。

import Foundation

struct requestbody: Codable {
    let data: DataClass
}

// MARK: - DataClass
struct DataClass: Codable {
    let slices: [Slice]
    let passengers: [Passenger]
    let cabinClass: String

    enum CodingKeys: String, CodingKey {
        case slices, passengers
        case cabinClass = "cabin_class"
    }
}

// MARK: - Passenger
struct Passenger: Codable {
    let type: String
}

// MARK: - Slice
struct Slice: Codable {
    let origin, destination, departureDate: String

    enum CodingKeys: String, CodingKey {
        case origin, destination
        case departureDate = "departure_date"
    }
}
class APIPost {

func getposts(response: requestbody) -> URLRequest? {
        guard let url = URL(string: "https://api.duffel.com/air/offer_requests") else { return nil }
        var request = URLRequest(url: url)
        request.setValue("application/json", forHTTPHeaderField: "Content-Type")
        request.setValue("application/json", forHTTPHeaderField: "Accept")
        request.setValue("beta", forHTTPHeaderField: "Duffel-Version")
        request.setValue("Bearer [redacted]", forHTTPHeaderField: "Authorization")
        request.httpMethod = "POST"
        request.httpBody = try! JSONEncoder().encode(response)
        return request
}

但是，我会说，当谈到如何集成此方法以解码 JSON 响应，然后在 UI 中使用时，我有些困惑。我也不能在函数中添加完成参数（添加@escaping），并且相信最好采取此操作并在另一个函数中构建它。任何高水平（或详细）的回复将不胜感激。

谢谢！

Answer 1

使用This generic function for calling Post方法，调用此方法并在完成时传递请求体和响应体，同样在

中添加URL

class func taskForPOSTRequest<RequestType: Encodable, ResponseType: Decodable>(url: URL, responseType: ResponseType.Type, body: RequestType, completion: @escaping (ResponseType?, Error?) -> Void) {
        var request = URLRequest(url: url)
        request.httpMethod = "POST"
        request.httpBody = try! JSONEncoder().encode(body)
        request.addValue("application/json", forHTTPHeaderField: "Content-Type")
        let task = URLSession.shared.dataTask(with: request) { data, response, error in
            guard let data = data else {
                DispatchQueue.main.async {
                    completion(nil, error)
                }
                return
            }
            let decoder = JSONDecoder()
            do {
                let responseObject = try decoder.decode(ResponseType.self, from: data)
                DispatchQueue.main.async {
                    completion(responseObject, nil)
                }
            } catch {
                print(error)
            }
        }
        task.resume()
    }

Answer 2

创建一个 URLRequest 实例是不够的，您还必须将此请求发送到您的服务器。您可以通过 URLSession API.

发送

// Notice the change in function signature
func getPost(requestBody: RequestBody, completion: @escaping ((_ posts: [Post]?, _ error: Error?) -> Void)) -> URLRequest? {
    
    // Everything else same up to this point
    request.httpBody = try! JSONEncoder().encode(requestBody)
    
    // Send the request to your server
    URLSession.shared.dataTask(with: request, completionHandler: { (data, urlResponse, error) in
       
        // Check if there was an error from network/server
        if let error = error {
            print(error.localizedDescription)
            // call completion with error from network/server
            completion(nil, error)
            return // return early because there was an error
        }
        
        // No error if we are, check for data
        if let data = data {
            do {
                // Try decoding your data to your model
                let posts = try JSONDecoder().decode([Post].self, from: data)
                // call completion with your models array, no error
                completion(posts, nil)
            }
            catch {
                // there's a decoding error, call completion with decoding error
                completion(nil, error)
            }
        }
    }).resume() // resuming a dataTask is essential, otherwise the call won't be sent to server
    
    return request
}

采用 Swift HTTP POST 并解码 JSON 响应

Taking Swift HTTP POST and Decoding JSON Response

api

xcode

httpresponse

func

swift