ios - 如何模拟 URL 结构？

Question

我正在编写一些代码以使测试更容易。经过研究，我发现使 URLSession 可测试的一个好方法是使其符合协议(protocol)并在我需要测试的类中使用该协议(protocol)。我对 URL 应用了相同的方法。但是，我现在需要将 url 参数向下转换为 URL 类型。这对我来说似乎有点“危险”。使 URL 可测试的正确方法是什么？我怎样才能模拟 dataTask 返回的 URLSessionDataTask 类型？

import Foundation

protocol URLSessionForApiRequest {
    func dataTask(with url: URL, completionHandler: @escaping (Data?, URLResponse?, Error?) -> Void) -> URLSessionDataTask
}

protocol URLForApiRequest {
    init?(string: String)
}

extension URLSession: URLSessionForApiRequest {}

extension URL: URLForApiRequest {}

class ApiRequest {
    class func makeRequest(url: URLForApiRequest, urlSession: URLSessionForApiRequest) {
        guard let url = url as? URL else { return }
        let task = urlSession.dataTask(with: url) { _, _, _ in print("DONE") }
        task.resume()
    }

}

Answer 1

您想要创建一个协议(protocol)来包装您要调用的函数，然后创建一个具体的实现和模拟实现，以返回您初始化它的任何内容。这是一个例子:

import UIKit
import PlaygroundSupport

protocol RequestProvider {
    func request(from: URL, completion: @escaping (Data?, URLResponse?, Error?) -> Void)
}

class ApiRequest: RequestProvider {
    func request(from url: URL, completion: @escaping (Data?, URLResponse?, Error?) -> Void) {
        URLSession.shared.dataTask(with: url, completionHandler: completion).resume
    }
}

class MockApiRequest: RequestProvider {
    enum Response {
        case value(Data, URLResponse), error(Error)
    }
    private let response: Response
    init(response: Response) {
        self.response = response
    }
    func request(from url: URL, completion: @escaping (Data?, URLResponse?, Error?) -> Void) {
        switch response {
        case .value(let data, let response):
            completion(data, response, nil)
        case .error(let error):
            completion(nil, nil, error)
        }
    }
}

class SomeClassThatMakesAnAPIRequest {
    private let requestProvider: RequestProvider
    init(requestProvider: RequestProvider) {
        self.requestProvider = requestProvider
    }
    //Use the requestProvider here and it uses either the Mock or the "real" API provider based don what you injected
}