iOS 9如何调用“返回"以编程方式进行功能 [英] iOS 9 how to invoke "Back to" feature programmatically

问题描述

是否有一个API可以通过编程方式调用返回"操作，以便返回到调用了(openurl)我的应用程序?

Is there an api to invoke "Back to" action programmatically in order to return to the app which called (openurl) mine?

推荐答案

您无法重定向回应用程序

It is not possible for you to redirect back to the app

原因是

以编程方式启动另一个应用程序(Apple公开)的唯一可能方法是通过URLScheme.这意味着当前应用程序应该知道将要启动的应用程序的URL方案. 在您的情况下，启动的应用程序不知道启动的应用程序的URL方案.因此，无法返回到先前的应用程序.

The only possible way for launching another app programmatically(Exposed by Apple) is via URLScheme. That means the current app should know the URL scheme of the to-be-launched app. In your case the launched app does not not know the URL scheme of the launched app. So there is no way to go back to the previous app.

这篇关于iOS 9如何调用“返回"以编程方式进行功能的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！