I'm writing an iOS app that allows other apps to open it via openURL.
Once done I want to send the user back where he came from without having him to press any button (including back).
How can I do this, if its possible ?
To be clear I want to maintain the status of the calling app. So using openURL again is not an option because I will lose context that way
Related
I have an audio app and I want to control play/pause with ApplicationShortcutItem directly, but I don't want to open the app.
Is it possible?
It is not possible to leave an app in the background when choosing a Home screen quick action.
From the documentation of UIApplicationShortcutItem:
When a user chooses one of your Home screen quick actions, the system launches or resumes your app and UIKit calls the application(_:performActionFor:completionHandler:) method in your app delegate.
There is a scenario where tapping on link open's iOS native browser after which I need to close the browser and switch back to applicatio. Is there any possible way we can do the using Selenium commands.
Consider using sendkeys method to tap at home button twice. Then try to tap at some place in the middle of the screen to get back to previously opened app. That could work, but I have not tried it.
#vignesh You can't switch between different apps in iOS. Currently, appium supports app switching between android apps only.
But you can run you app in background and after specified time it will be foreground.
You can use the following code to move back from the browser.
driver.navigate().back();
Just try this out.
I'm developing an iOS 7+ app that I need to offer the option of navigating to a certain web page to let the users to fill in a form there, and after that to come back to the app's view where the user was.
Is it possible to programmatically open Safari with a given url? If it is, I suppose that then there is no way to automatically redirect the user to your app from there... right? Is then a UIWebView the only option? Is it possible to navigate back or dismiss the view with the UIWebView without the need of user interaction?
Thanks
You can open links in Safari as detailed in this post How to launch safari and open URL from iOS app
I don't believe you can set a 'callback' and have it return to your app on completion, as you have no control over the user once they have exited your app's sandbox.
Opening the link in UIWebView would provide control, as you can utilize the UIWebView callbacks.
When I create an app using this component it just send notifications to Android while APP is opened or when she's in the background because I pressed the home button on the phone, if I close the app or exit using the back button notifications are no longer passed to the cell. Is there any way to fix this?
How do you expect for your app to create any notifications if it is not running? Your app needs to be rinning in order to create or post any notivications to the OS.
So now you would probably wanna focus on preventing the app instantly closing when back button is pressed in a way that you notiffy the user that closing the app will prevent its notifications to work and then give your user choice to either close the app or simply keep it workinng in the background.
My question:
I have an iOS app, it is minimized, and I want it to start when user press home button three times (or, for example combination of buttons, like "HOME+Volume-" and so on). Can I make it for non-jailbroken devices, only with native iOS functions?
It's not possible to customise iOS like that.
Without jailbraking, apps can only be opened by using custom URL schemes from other apps or by normally tapping on the on the home screen. To have an app open like this you would have to modify the OS and there is no way you can do that without a jailbreak.
Standard OS functionality like pressing the home button multiple times cannot be overridden because Apple is really against modifying any part of way the user interacts with apps or the OS.
No.
Not possible without jailbreak and then it would have to be some sort of preference in the OS settings.