I am using the below code to dial a phone number that I get from contacts:
func dialNumber(number : String) {
let formatedNumber = number.components(separatedBy: NSCharacterSet.decimalDigits.inverted).joined(separator: "")
print("formatedNumber:", formatedNumber)
if let url = URL(string: "tel://\(formatedNumber)"),
UIApplication.shared.canOpenURL(url) {
if #available(iOS 10, *) {
UIApplication.shared.open(url, options: [:], completionHandler:nil)
} else {
UIApplication.shared.openURL(url)
}
}
}
Additional info:
var urlTest = URL(string: "tel:+12345678900")!
print("UIApplication.shared.canOpenURL(url):", UIApplication.shared.canOpenURL(urlTest)) // Returns true
UIApplication.shared.open(urlTest, options: [:], completionHandler:nil)
var urlTest = URL(string: "tel:2345678900")!
print("UIApplication.shared.canOpenURL(url):", UIApplication.shared.canOpenURL(urlTest)) // Returns true
UIApplication.shared.open(urlTest, options: [:], completionHandler:nil)
The code works for numbers with Area code +1XXXYYYZZZZ and the phone number is dialed.
But it doesn't work for numbers like XXXYYYZZZZ.
If I open the contact using Address Book, the number XXXYYYZZZZ is recognized as a phone number. On clicking, the number gets dialed.
How can I get a similar behaviour as Address Book inside my app?
I do not want to add a region code +1 as the app will be used in other countries as well.
There are external libraries for identifying phone numbers but I'm trying to explore if there is a native solution.
If a number is not recognized, I even tried to open the contact in the address book so the user can dial the number from the contacts app, but it looks like address book is not a supported scheme.
I have a problem. I am trying to create call by phone number function. But I need to have a phone number in the format (###)###-#### at standard iOS alert. And for some reason, this function gives different results on different screens. How to change this?
if let phoneCallURL:URL = URL(string: "tel://\(phoneNumber)") {
let application:UIApplication = UIApplication.shared
if (application.canOpenURL(phoneCallURL)) {
if #available(iOS 10, *) {
application.open(phoneCallURL, options: [:], completionHandler: {success in
print("Open tel: \(success)")
})
} else {
let success = UIApplication.shared.openURL(phoneCallURL)
print("Open: \(success)")
}
}
}
Page 1:
Page 2:
Update:
I tried use formatted strings, but at first page I get unformatted phone number, and at second page I get phone number with this format: #(##) ### ## ##. Always. No matter I use formatted or unformatted phone number.
It seems to me what must be some kind of extension for managing the telephone label in the system alert.
did anyone successfully deep linked from his app to a directions to a place in moovit ? why can't i ?! it just opens the app and does nothing ....
if anyone successfully deep linked to a direction to anywhere please help.
if UIApplication.shared.canOpenURL(URL(string: "moovit://")!) {
// Moovit installed - launch app (with parameters)
let MoovitURL: String = "moovit://directions?dest_lat=40.758896&dest_lon=-73.985130&dest_name=TimesSquare&orig_lat=40.735845&orig_lon=-73.990512&orig_name=UnionSquare&auto_run=true&partner_id=<testApp2345>"
let escapedString = MoovitURL.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
UIApplication.shared.openURL(URL(string: escapedString!)!)
}else {
// Moovit not installed - send to store
UIApplication.shared.openURL(URL(string: "https://itunes.apple.com/us/app/id498477945")!)
}
}
even Moovit's own example isn't working for me....Whats wrong?
The problem is with the .urlHostAllowed parameter.
Using the .urlQueryAllowed parameter instead will only convert the parameters after '?'.
Fixed your code:
if UIApplication.shared.canOpenURL(URL(string: "moovit://")!) {
// Moovit installed - launch app (with parameters)
let MoovitURL: String = "moovit://directions?dest_lat=40.758896&dest_lon=-73.985130&dest_name=TimesSquare&orig_lat=40.735845&orig_lon=-73.990512&orig_name=UnionSquare&auto_run=true&partner_id=<testApp2345>"
let escapedString = MoovitURL.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
UIApplication.shared.open(URL(string: escapedString!)!, options: [:], completionHandler: nil)
}else {
// Moovit not installed - send to store
UIApplication.shared.open(URL(string: "https://itunes.apple.com/us/app/id498477945")!, options: [:], completionHandler: nil)
}
Try opening this url in Safari and see if it works (It does for me).
Looks like the problem is with the code, maybe try without addingPercentEncoding or use the new openURL API
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")
}
what my app currently has is when user tapped the link it opens the app and it works perfectly but my problem is I add something to the prefix of my url
this is the original url to be catch in my app
www.mywebsiteurl.com?variable=catchthisvalue
I achieved this by using URL Schemes and this is the prefix I added to my link myappname:// and the link become this
myappname://www.mywebsiteurl.com?variable=catchthisvalue
to get the original link I added this code to my appDelegate
func application(application: UIApplication, openURL url: NSURL, sourceApplication: String?, annotation: AnyObject) -> Bool {
let cleanString = url.absoluteString.stringByReplacingOccurrencesOfString("myappname://", withString: "")
BaseURL.shared.isSet = String(cleanString)
return true
}
the cleanString is the original url I saved it to NSUserDefaults, this process is perfectly working, but my client don't want to change the url
my question is can I open my app without adding prefix or anything in original url?