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63 lines
2.6 KiB
Markdown
63 lines
2.6 KiB
Markdown
# 创建自定义的插件
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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)
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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:
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```swift
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final class RequestAlertPlugin: PluginType {
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private let viewController: UIViewController
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init(viewController: UIViewController) {
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self.viewController = viewController
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}
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func willSend(_ request: RequestType, target: TargetType) {
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}
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func didReceive(_ result: Result<Moya.Response, MoyaError>, target: TargetType) {
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}
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}
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```
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然后，当请求将被发送时，我们向函数添加一些功能:
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```swift
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func willSend(_ request: RequestType, target: TargetType) {
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//make sure we have a URL string to display
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guard let requestURLString = request.request?.url?.absoluteString else { return }
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//create alert view controller with a single action
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let alertViewController = UIAlertController(title: "Sending Request", message: requestURLString, preferredStyle: .alert)
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alertViewController.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
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//and present using the view controller we created at initialization
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viewController.present(viewControllerToPresent: alertViewController, animated: true)
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}
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```
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最后, 如果结果出错了，让我们在 `didReceive` 中实现一个提示框
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```swift
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func didReceive(_ result: Result<Moya.Response, MoyaError>, target: TargetType) {
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//only continue if result is a failure
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guard case Result.failure(_) = result else { return }
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//create alert view controller with a single action and messing displaying status code
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let alertViewController = UIAlertController(title: "Error", message: "Request failed with status code: \(error.response?.statusCode ?? 0)", preferredStyle: .alert)
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alertViewController.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
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//and present using the view controller we created at initialization
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viewController.present(viewControllerToPresent: alertViewController, animated: true)
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}
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```
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到此完毕, you now have very well informed, if slightly annoyed users.
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_(Please note that this example will usually fail since presenting an alert twice on the same view controller is not allowed)_
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