I wanna go to settings if the user clicks on "go to settings", can someone please help?
heres what the notification looks like
Try this:
if let appSettings = NSURL(string: UIApplicationOpenSettingsURLString) {
UIApplication.sharedApplication().openURL(appSettings)
}
Related
I have a Share-Extension from which I would like to redirect the user to the main app. This is what I tried:
let signInAction = UIAlertAction(title: "Anmelden", style: .default) { (alert) in
let myAppUrl = NSURL(string: "open://")!
self.extensionContext?.open(myAppUrl as URL, completionHandler: { (success) in
if (!success) {
// let the user know it failed
print("fail!")
}
})
}
And my info-plist :
Right now it is printing the "fail!". What am I doing wrong here?
Have a look at your project settings in xcode.
Did you set a custom url scheme in the "URL-Types" setting in your apps target?
Xcode - Target - General Settings - URL Types
// 1
let urlWhats = "https://wa.me/\(mobile)/?text=\(text)"
// 2
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed) {
// 3
if let whatsappURL = NSURL(string: urlString) {
// 4
if UIApplication.shared.canOpenURL(whatsappURL as URL) {
// 5
UIApplication.shared.open(whatsappURL as URL, options: [:], completionHandler: nil)
// UIApplication.shared.\
} else {
// 6
print("Cannot Open Whatsapp")
}
}
}
I'm able to launch whatsapp from my app from the above mentioned code, it is composing prefix text to the contact I wish to send and I need to click the send button in whatsapp manually . But I'm looking for a code which automatically sends whatsapp text to number from my app. Can anyone share your thoughts on this?
You can only compose the message for a particular contact using the Deep Linking method that you have used for it. For sending the message user has to click on the send button manually. You could provide the user with an alert that says so. But, it's not possible to do it for the user from your side. If you were able to send a message on Whatsapp by writing code without the user's confirmation it would be a break of user's privacy. Don't you think?
I know there are lot of topics about this problem but is there a solution? I have not found official documentation for that.
My problem is I need to redirect my code on storage settings like this : (work in iOS 9)
let settingsUrl = NSURL(string: "prefs:root=CASTLE&path=STORAGE_AND_BACKUP")
if let url = settingsUrl {
UIApplication.shared.openURL(url as URL)
}
But since ios10 this method don't work, so Is there an alternative? I saw SnapChat, Google Maps redirect their apps to different part on settings (not the main screen of settings) so I think there is a solution.
I already implement the url scheme in info.plist but it's still not work
I tried this method but same issue too
let settingsUrl = NSURL(string: "prefs:root=CASTLE&path=STORAGE_AND_BACKUP")
if let url = settingsUrl {
if #available(iOS 10.0, *) {
UIApplication.shared.open(URL.init(string:"Prefs:root=General&path=STORAGE_ICLOUD_USAGE/DEVICE_STORAGE")!, options: [UIApplicationOpenURLOptionUniversalLinksOnly:true], completionHandler:{(success: Bool?) -> Void in})
} else {
UIApplication.shared.openURL(url as URL)
}
}
Thank in advance.
Since iOS 10, it's not possible to open the Settings app from a third party app. The only settings that are allowed to be opened are Keyboard setting but only by a custom keyboard extension and your own application settings.
More details: here
Note: Even for iOS 9, using the URL string that is mentioned in the question can lead to app rejection as it violates iOS App Review Guidelines.
So I am about to launch an app to the App Store. My issue is that I have a rate my app please button but I do not know the right code to insert there.
My fiends tried this on their app and said it was no good:
Does anybody know how I can fix this issue?
let iTunesReviewPageLink = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?id=1073785561&pageNumber=0&sortOrdering=2&type=Purple+Software&mt=8"
// Go directly to review page on App Store
if let url = NSURL(string: iTunesReviewPageLink) {
UIApplication.sharedApplication().openURL(url)
}
The only unknown thing is the ID, right? You can see the ID of your app before it is published - once you set it up in iTunes Connect.
If your app is not yet released, you haven't got an App Store link. So that's impossible.
In order to implement this functionality when your app is released, you can use the following code:
Swift 2
let appIDString = "APP_ID" // your app ID
let reviewsURLString = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?pageNumber=0&sortOrdering=1&type=Purple+Software&mt=8&id=\(appIDString)"
let reviewsURL = NSURL(string: reviewsURLString)
if reviewsURL != nil && UIApplication.sharedApplication().canOpenURL(reviewsURL!) {
UIApplication.sharedApplication().openURL(reviewsURL!)
}
else {
// handle situation if reviews url cannot be opened.
}
Swift 3
let appIDString = "APP_ID" // your app ID
let reviewsURLString = "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?pageNumber=0&sortOrdering=1&type=Purple+Software&mt=8&id=\(appIDString)"
let reviewsURL = URL(string: reviewsURLString)
if reviewsURL != nil && UIApplication.shared.canOpenURL(reviewsURL!) {
UIApplication.shared.openURL(reviewsURL!)
}
else {
// handle situation if reviews url cannot be opened.
}
EDIT:
This links works in iOS 8 and 9 and links directly to reviews page of the app in App Store application.
I am not sure about iOS 7. Probably for iOS 7 you need to use different link.
I'm trying to open whatsapp conversation/chat for particular contact.
Instead of opening the desired chat it only opens the app.
No idea whats wrong :
let URLString = "whatsapp://send?abid=\(ID);text=lOL;"
UIApplication.sharedApplication().openURL(NSURL(string: URLString)!)
URLString value : whatsapp://send?abid=414;text=lOL
Any suggestions?
Update your URL like this:
whatsapp://send?abid=\(ID)&text=lOL
Source from HERE.
Try this and check if the UIApplication and open the URL.
let whatsAppURL: NSURL = NSURL(string: "whatsapp://send?abid=\(ID)&text=lOL")
if UIApplication.sharedApplication().canOpenURL(whatsAppURL){
UIApplication.sharedApplication().openURL(whatsAppURL)
}