swift - MVC 模型中的动态链接

Question

在此示例中，如何随时发送动态链接到服务类，在此更新视图之后，我随时收到链接，我不想使用通知等

1. 看法

   var viewModel = HomeViewModel()
   
    func configureViewModel() {
        viewModel.resultClosure = { [weak self] listModel in
            guard let image = listModel.uiImage else { return }
            self?.updateImageViewConstraint(nil, image: image)
        }
    }
   

2. HomveViewModel

   class HomeViewModel {
    var service = Services()
    var listModel = [ListModel]() {
           didSet {
               DispatchQueue.main.async { [weak self] in
                   guard let self = self else {
                       return
                   }
                    self.resultClosure?(self.listModel[0])
               }
           }
       }
   
    let placeholderImage = UIImage(named: "placeholder")
    var updatedImge : UIImage?
    var resultClosure: ((ListModel) -> Void)?
    var stringUrl: StringUrl!
   
    func fetchData() {
        service.makeRequest() { (result: Result<[ListModel]>) in
            switch result {
            case .success(let listModel):
                self.listModel = listModel                    
                return
            case .failure(let error):
                print("\n HomeViewModel fetchData failure \(error) \n")
                return
            }
        }
    }
   

   }

3. 服务 - 这里我需要检查动态链接

   class Services {
    let urlSession: URLSession
    let imageProvider = ImageProvider()
    let dynamicLinkfromStruct = "Dynamic Link From Struct!!!" 
   
   
    init(urlSession: URLSession = .shared) {
        self.urlSession = urlSession
    }
   
   
    func makeRequest(completionHandler: @escaping (Result<[ListModel]>) -> Void) {
   
   
    guard let url = URL(string: dynamicLinkfromStruct) else { return } 
   
        imageProvider.loadImages(from: url) { (image) in
            do {
                let result = [ListModel.init(imageUrl: self.url, uiImage: image)] 
                completionHandler(.success(result))
            } catch {
                completionHandler(.failure(error ))
            }
        }
    }
   }
   

4. 模型

   struct ListModel {
   
    let imageUrl: String?
    let uiImage: UIImage?
   }
   

5. 构造我需要在视图中显示的任何时间动态链接。在这里提问！如何发送服务链接和更新视图？？？

   struct GetManager {
    static func showContent(pasteboard: String) {
    switch onlyLinkAbsoluteString.validURL {
           case true:
    // How to Send Link to Services and update View???
    }
    }
    }
   

标签： swiftmodel-view-controllerstruct

Answer 1

在 MVVM/MVC 通信中应该走这条路

视图 > ViewModel > 管理器 > 服务

在 ViewModel 中创建 Manager 实例，在 Manager 中创建 Service 实例

您在 ViewModel 中调用 Manager 的函数，Manager 将处理与您的 Service 的通信。