I want to open some URLs with Safari browser of apple device. And my openURL function works fine most of the time. However when I tried to open this URL, it failed with the following error.
Download Failed Safari cannot download file.
Here is the URL that fails:
https://www.youtube.com/v/QH2-TGUlwu4?version=3&autohide=1
Here is my code:
let url = "https://www.youtube.com/v/QH2-TGUlwu4?version=3&autohide=1"
if let nsurl = NSURL(string: url){
if UIApplication.sharedApplication().canOpenURL(nsurl){
UIApplication.sharedApplication().openURL(nsurl)
}else{
print("Cannot open this NSURL.")
}
}else{
print("Cannot convert String to NSURL.")
}
The URL you mentioned
https://www.youtube.com/v/QH2-TGUlwu4?version=3&autohide=1
contains a SWF file i.e. Shockwave Flash file.
Flash cannot played on an iOS based device (link).
1) Either you need to show appropriate error message to user.
2) Or you can redirect user to YouTube app (instead of Safari), where user would be able stream and watch the video.
See this Apple doc link to understand how to redirect user to YouTube app.
You need to encode URL before use it like:
let urlString = "https://www.youtube.com/v/QH2-TGUlwu4?version=3&autohide=1"
var escapedString = urlString.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet())
if let nsurl = NSURL(string: escapedString){
if UIApplication.sharedApplication().canOpenURL(nsurl){
UIApplication.sharedApplication().openURL(nsurl)
}else{
print("Cannot open this NSURL.")
}
}else{
print("Cannot convert String to NSURL.")
}
I don't know how to check if app is installed or not on phone! Or when App is installed, open the app, otherwise open the Appstore link to download the app. I'm using swift 3. I want to do it using app name or bundle identifier.
Thank You!
func openApp(appName:String) {
let appName = "instagram"
let appScheme = "\(appName)://app"
let appUrl = URL(string: appScheme)
if UIApplication.shared.canOpenURL(appUrl! as URL) {
UIApplication.shared.open(appUrl!)
} else {
print("App not installed")
}
}
After looking into so many answers, i am writing a common code which will help for new users. If you have two mobile apps as App1 and App2, if you want to check in App2 that App1 is already installed in his device or not, here is code below.
In App1 add this property in info.plist file
<key>CFBundleURLTypes</key>
<array>
<dict>
<key>CFBundleURLName</key>
<string>com.companyName.App1</string>
<key>CFBundleURLSchemes</key>
<array>
<string>App1</string>
</array>
</dict>
</array>
In App2 add this property in info.plist file
<key>LSApplicationQueriesSchemes</key>
<array>
<string>App1</string>
</array>
In App2 write the method to check if app is installed or not on phone! Or when App is installed, open the app, otherwise open the Appstore link to download the app as below.
func checkAndOpenApp(){
let app = UIApplication.shared
let appScheme = "App1://app"
if app.canOpenURL(URL(string: appScheme)!) {
print("App is install and can be opened")
let url = URL(string:appScheme)!
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
} else {
print("App in not installed. Go to AppStore")
if let url = URL(string: "https://apps.apple.com/us/app/App1/id1445847940?ls=1"),
UIApplication.shared.canOpenURL(url)
{
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
}
}
}
I hope it will help some one.
Between the other answers and their comments, I'm getting the impression that the asker wants to be able to see if any given app is installed.
Beginning with iOS 9.0, that is not possible.
Apps for iOS 9 and later must have a list of requested URL schemes in the Info.plist before being allowed to use canOpenURL:. This is to protect user privacy, as advertisers were abusing this functionality in an arguably invasive fashion. (See this excellent post for more details on those changes.)
Of course, that list is static and cannot be changed after build time or submission to the App Store. If Apple doesn't like the ones you chose, they have every right to reject it.
I'm afraid that what you're asking isn't possible within reason for iOS 9.0 and later.
Edit: I also want to make clear that an app's URL scheme may not necessarily match nicely with its name. (This is more of an issue of a badly named constant than a functional issue.) There used to be a giant list of known URI schemes with documentation for each, but poignantly and fittingly enough, it seems that the wiki hosting it has shut down.
Swift 4.1. One developer can have more than one app on AppStore. So, I need to check if user has installed other apps or not by the same developer. I had Bundle ID's of other apps. Although you can use Appname instead of Bundle Id. So I followed the following steps.
In your current app add LSApplicationQueriesSchemes key of type Array in your info.plist file. Make entry of bundle id or Appname of app there which you want to open from your app.
Other app should have their bundle id or Appname entry in that app URL Scheme.
In your current app check if that app in installed in iPhone or not and can open accordingly.
let app = UIApplication.shared
let bundleID = "some.Bundle.Id://"
if app.canOpenURL(URL(string: bundleID)!) {
print("App is install and can be opened")
let url = URL(string:bundleID)!
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
} else {
print("App in not installed. Go to AppStore")
}
You can also test it from Safari browser. Just type the following in search bar
URL_scheme:// or Bundle_Id://
If app is installed the it will show alert with Appname to open it in app.
This worked for me (Swift 3.0)
Below two inputs should be provided:
<APP URL SCEHME>: The URL Scheme of the app which you want to open
<YOUR APP URL>: The App Itunes URL
func openApp() {
let appURL = NSURL(string: "<APP URL SCHEME>")
if UIApplication.shared.canOpenURL(appURL as! URL) {
print("Opening App...")
}else {
UIApplication.shared.openURL(NSURL(string: "<YOUR APP URL>")! as URL)
}
}
first go to info.plist, add LSApplicationQueriesSchemes add an item and place instagram on that item. Now this code will run perfectly.
let appName = "instagram"
let appScheme = "\(appName)://"
let appUrl = URL(string: appScheme)
if UIApplication.shared.canOpenURL(appUrl! as URL)
{
UIApplication.shared.open(appUrl!)
} else {
print("App not installed")
}
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.
Try as I might I cannot get a simple app to dial a number.
I hooked a button action to this function:
#IBAction func wackAMole(sender: AnyObject) {
var phoneNumber = "tel://555-555-5555"
var url = NSURL(fileURLWithPath: phoneNumber)
var sharedApplication = UIApplication.sharedApplication()
if (sharedApplication.canOpenURL(url)) {
sharedApplication.openURL(url)
}
}
I have also added "telephony" to Info.plist as a "Required device capabilities" Still to no avail. When running under the debugger on an iPhone 5 canOpenURL() returns false.
Even if I change phoneNumber to "https://google.com", canOpenURL() will return false.
Any ideas where to look next?
You aren't creating a URL for a local file system entry, so you shouldn't use fileURLwithPath. You should use URLWithString -
var url = NSURL.URLWithString(phoneNumber)