docs_CN/Examples/CustomPlugin.md
作为构建一个自定义的插件的示例，假设我们想把网络活动通知给用户，那么当请求被发送时，我们将显示携有关于请求的基本信息的提示框，并且当一个响应表明这个请求失败时让用户知道 (对于这个例子，我们还需要假设我们不介意用大量的提示框来打扰用户)
首先，我们需要创建一个遵循 PluginType的类, 并且它接收一个视图控制器 (用来展示 UIAlertController)的实例对象的引用:
final class RequestAlertPlugin: PluginType {
private let viewController: UIViewController
init(viewController: UIViewController) {
self.viewController = viewController
}
func willSend(_ request: RequestType, target: TargetType) {
}
func didReceive(_ result: Result<Moya.Response, MoyaError>, target: TargetType) {
}
}
然后，当请求将被发送时，我们向函数添加一些功能:
func willSend(_ request: RequestType, target: TargetType) {
//make sure we have a URL string to display
guard let requestURLString = request.request?.url?.absoluteString else { return }
//create alert view controller with a single action
let alertViewController = UIAlertController(title: "Sending Request", message: requestURLString, preferredStyle: .alert)
alertViewController.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
//and present using the view controller we created at initialization
viewController.present(viewControllerToPresent: alertViewController, animated: true)
}
最后, 如果结果出错了，让我们在 didReceive 中实现一个提示框
func didReceive(_ result: Result<Moya.Response, MoyaError>, target: TargetType) {
//only continue if result is a failure
guard case Result.failure(_) = result else { return }
//create alert view controller with a single action and messing displaying status code
let alertViewController = UIAlertController(title: "Error", message: "Request failed with status code: \(error.response?.statusCode ?? 0)", preferredStyle: .alert)
alertViewController.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
//and present using the view controller we created at initialization
viewController.present(viewControllerToPresent: alertViewController, animated: true)
}
到此完毕, you now have very well informed, if slightly annoyed users.
(Please note that this example will usually fail since presenting an alert twice on the same view controller is not allowed)