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iphone:如何为用户每 5 次打开应用程序时对 iphone 应用程序的应用程序弹出率进行评分?

[英]iphone:how to implement rate the app popup for iphone app every 5th time user opens up app?

I need to implement the rate tha app popup when user every 5th time open the app user can also give review当用户每 5 次打开应用程序时,我需要实现应用程序弹出率,用户也可以给予评论

and then i need to redirect him to page of iTunes然后我需要将他重定向到 iTunes 页面

Since iOS 10.3+ the right way to request app reviews is through SKStoreReviewController, which is Apple's official way of requesting App Store ratings and reviews from users.自 iOS 10.3+ 起,请求应用程序评论的正确方法是通过 SKStoreReviewController,这是 Apple 向用户请求 App Store 评分和评论的官方方式。

Simply call:只需调用:

[SKStoreReviewController requestReview];

More details on the class:SKStoreReviewController有关 class 的更多详细信息:SKStoreReviewController

Have a look at the excellent Appirater library.看看优秀的Appirater库。 The APPIRATER_USES_UNTIL_PROMPT constant in Appirater.h defines often the prompt appears. Appirater.h 中的APPIRATER_USES_UNTIL_PROMPT Appirater.h定义了经常出现的提示。 The prompt reappears in new versions of your app!提示会重新出现在您的应用程序的新版本中!

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