The problem I have is that I want to have a settings button which when the user clicks it, takes them to a specified path in the settings such as sound settings. The only issue is that every method I've tried has been outdated for more recent versions of iOS 10. Does anyone have any suggestions for a method which I can use? Any help will be greatly appreciated.
You can open your app Settings page by...
let settings_app: URL = URL(string: UIApplicationOpenSettingsURLString)!
UIApplication.shared.open(settings_app)
But you can't open other Settings app section from any app.
Try the following code, I hope to help you
let settingsUrl = NSURL(string:UIApplicationOpenSettingsURLString) as! URL
if #available(iOS 10.0, *) {
UIApplication.shared.open(settingsUrl, options: [:], completionHandler: nil)
} else {
// Fallback on earlier versions
}
Related
I am using Xcode 11.3.1 and try to login with snapchat with loginkit I add the add information in info.plist and my code is
SCSDKLoginClient.login(from: self, completion: { success, error in
if let error = error {
print(error.localizedDescription)
return
}
if success {
self.fetchSnapUserInfo() //example code
}
})
this code show me the login ui of snapchat and I am login into snapchat with my account.
but I am stuck on this ui
when I am click on continue nothing is happing . SCSDKLoginClient completion block not called.
Please ensure your URL scheme is configured correctly.
URL Schemes
hello everyOne soo finally i found the solution
i am using 11.3.1 and when i create new project the add
AppDelegate and SceneDelegate default class.
so according to snapchat logkit documentation i add
func application(_ app: UIApplication,
open url: URL,
options: [UIApplication.OpenURLOptionsKey : Any] = [:]) -> Bool {
return SCSDKLoginClient.application(app, open: url, options: options)
}
method in my Appdelegate class. but this method never get called in xocode 11.3.1
so the solution of my problem is this
func scene(_ scene: UIScene, openURLContexts URLContexts: Set<UIOpenURLContext>) {
guard let url = URLContexts.first?.url else {
return
}
SCSDKLoginClient.application( UIApplication.shared, open: url, options: nil)
}
you need to add this method into your sceneDelegate file. then everything work fine.
synpchat need to update there doc for new xcode 11.3.1. i hope this answer help you guy's because i also wast my 3 day's on this issue.
happy Coding :)
In my iOS app I have a bunch of mp4 videos that I download at a certain time on the app using On Demand Resources. Using this tutorial:
https://www.raywenderlich.com/520-on-demand-resources-in-ios-tutorial
I download the resources like this at the start of the app, in a previous view controller:
func requestSceneWith(tag: String,
onSuccess: #escaping () -> Void,
onFailure: #escaping (NSError) -> Void) {
// 2
currentRequest = NSBundleResourceRequest(tags: [tag])
// 3
guard let request = currentRequest else { return }
request.beginAccessingResources { (error: Error?) in
// 4
if let error = error {
onFailure(error as NSError)
return
}
// 5
onSuccess()
}
The resource seem to download fine, and I know that they have been downloaded, by looking in the disk report in xcode.
However, when the videos are supposed to be played in the app, the app just shows a black screen. Here is my code to play the videos:
let videoURL = Bundle.main.url(forResource: "cow2", withExtension: "mp4", subdirectory: "Videos/Animals")
self.player = AVPlayer(url: videoURL!)
self.myPlayerController.player = self.player
self.myPlayerController.player?.play()
Now, when the resources are not tagged, and they come with the app and not downloaded later, they work fine. And the console prints the file name, like I did (print(videoURL.absoluteString). But after they are tagged and downloaded later, they dont work, and nothing prints in the console. Just a black screen appears in the app.
I've been stuck on this for ages, and help with really help.
Thanks
I think you didn't request the on-demand resources before accessing the video file. Maybe you can try to declare the NSBundleResourceRequest instance as a global variable in AppDelegate.
This question already has an answer here:
ios User Prompt when making outgoing call via URL Scheme 10.2+
(1 answer)
Closed 4 years ago.
for my company, I need my app to make periodically some phone call to a given number and then analyze call quality.
I'm testing
private func callNumber(phoneNumber:String) {
if let phoneCallURL = URL(string: "telprompt://\(phoneNumber)") {
let application:UIApplication = UIApplication.shared
if (application.canOpenURL(phoneCallURL)) {
if #available(iOS 10.0, *) {
application.open(phoneCallURL, options: [:], completionHandler: nil)
} else {
// Fallback on earlier versions
UIApplication.shared.openURL(phoneCallURL as URL)
}
}
}
}
but this code requires for a an "ok" on confirmation dialog. How can I avoid it? I found some answer, but they are old and in objective c
Taken from the iOS SDK Release Notes for iOS 10.3 https://developer.apple.com/library/archive/releasenotes/General/RN-iOSSDK-10.3/
openURL
When a third party application invokes openURL: on a tel://, facetime://, or facetime-audio:// URL, iOS displays a prompt and requires user confirmation before dialing.
I'm working on a macOS cocoa-app in Swift where I import several different file types into the app for the user to interact with.
I'm currently trying to determine if it's possible to implement the "Open file with" feature, so that the user could open those files in a different program if they wanted to:
I've found a few different SO questions that seem tangentially related to what I'm trying to do:
Swift: How to open file with associated application?
Launch OSX Finder window with specific files selected
...but so far nothing to indicate if it's possible to implement right-click Finder/file (?) access in the way I had in mind.
Apologies if this is too vague of a question; any help / guidance appreciated!
Without going into details, it's pretty straight forward:
Get the list of all known applications that can open a specific file type (see LSCopyApplicationURLsForURL, a Core Foundation C function).
Build the menu. You can use NSWorkspace (and probably URL) to get the application icons.
Use NSWorkspace.openFile(_:withApplication:) to tell the application to open the given document.
2022, Swift 5
Get app list associated with local file:
func getAppsAssociatedWith(_ url: URL?) {
guard let url = localFileURL,
let retainedArr = LSCopyApplicationURLsForURL( url as CFURL, .all)?.takeRetainedValue(),
let listOfRelatedApps = retainedArr as? Array<URL>
else {
return []
}
return listOfRelatedApps
}
Getting thumbnail for app:
let singleAppIcon = NSWorkspace.shared
.icon(forFile: appUrl.path)
.scaledCopy(sizeOfLargerSide: 17)
Open url with app:
#available(macOS 10.15, iOS 9.0, *)
public class func openUrlWithApp(_ urls: [URL], appUrl: URL) {
NSWorkspace.shared.open(urls, withApplicationAt: appUrl, configuration: NSWorkspace.OpenConfiguration())
}
In my app I'm cashing all apps icons in dictionary.
[someFile localURL : app icon]
If I have already got icon earlier - no need to get it once more
var relatedAppsThumbnails: [URL: Image] = [:]
func updateRelatedApps() {
guard let url = currImgUrl, // file url to get icons from related apps
let retainedArr = LSCopyApplicationURLsForURL( url as CFURL, .all)?.takeRetainedValue(),
let listOfRelatedApps = retainedArr as? Array<URL>
else {
relatedApps = []
return
}
self.relatedApps = listOfRelatedApps
// add app icon in case of it wasn't added yet
for appUrl in listOfRelatedApps {
if relatedAppsThumbnails[appUrl] == nil {
let nsImg = NSWorkspace.shared.icon(forFile: appUrl.path)
.scaledCopy(sizeOfLargerSide: 17)
relatedAppsThumbnails[appUrl] = Image(nsImage: nsImg)
}
}
}
LSCopyApplicationURLsForURL is deprecated. You can use this alternative:
func getListOfExternalApps(forURL url: URL) -> [(URL, Image)] {
let listOfExternalApps = NSWorkspace.shared.urlsForApplications(toOpen: url)
let icons = listOfExternalApps.map {
let nsimage = NSWorkspace.shared.icon(forFile: $0.path())
nsimage.size = CGSize(width: .s16, height: .s16)
return Image(nsImage: nsimage)
}
return Array(zip(listOfExternalApps, icons))
}
I have spent some days to find the Swift code for youtube Channel to open from my app. but i couldn't find At all, can some one help me please!!
i need the Code in Swift.
Update for Swift 3 and iOS 10+
OK, here is how to do it in Swift 3. Basically, there are two easy steps to achieve this:
First, you have to modify Info.plist to list Youtube with LSApplicationQueriesSchemes. Simply open Info.plist as a Source Code, and paste this:
<key>LSApplicationQueriesSchemes</key>
<array>
<string>youtube</string>
</array>
After that, you can open any youtube URL inside Youtube application by simply substituting https:// with youtube://. Here is a complete code, you can link this code to any button you have as an Action:
#IBAction func YoutubeAction() {
let YoutubeUser = "Your Username"
let appURL = NSURL(string: "youtube://www.youtube.com/user/\(YoutubeUser)")!
let webURL = NSURL(string: "https://www.youtube.com/user/\(YoutubeUser)")!
let application = UIApplication.shared
if application.canOpenURL(appURL as URL) {
application.open(appURL as URL)
} else {
// if Youtube app is not installed, open URL inside Safari
application.open(webURL as URL)
}
}
I hope this suits your purpose:
var url = NSURL(string: "https://youtube.com/your-channel")
if UIApplication.sharedApplication().canOpenURL(url!) == true {
UIApplication.sharedApplication().openURL(url!)
}