Make a phone call programmatically

Merging the answers of @Cristian Radu and @Craig Mellon, and the comment from @joel.d, you should do:

NSURL *urlOption1 = [NSURL URLWithString:[@"telprompt://" stringByAppendingString:phone]];
NSURL *urlOption2 = [NSURL URLWithString:[@"tel://" stringByAppendingString:phone]];
NSURL *targetURL = nil;

if ([UIApplication.sharedApplication canOpenURL:urlOption1]) {
targetURL = urlOption1;
} else if ([UIApplication.sharedApplication canOpenURL:urlOption2]) {
targetURL = urlOption2;
}

if (targetURL) {
if (@available(iOS 10.0, *)) {
[UIApplication.sharedApplication openURL:targetURL options:@{} completionHandler:nil];
} else {
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wdeprecated-declarations"
[UIApplication.sharedApplication openURL:targetURL];
#pragma clang diagnostic pop
}
}

This will first try to use the "telprompt://" URL, and if that fails, it will use the "tel://" URL. If both fails, you're trying to place a phone call on an iPad or iPod Touch.

Swift Version :

let phone = mymobileNO.titleLabel.text
let phoneUrl = URL(string: "telprompt://\(phone)"
let phoneFallbackUrl = URL(string: "tel://\(phone)"
if(phoneUrl != nil && UIApplication.shared.canOpenUrl(phoneUrl!)) {
UIApplication.shared.open(phoneUrl!, options:[String:Any]()) { (success) in
if(!success) {
// Show an error message: Failed opening the url
}
}
} else if(phoneFallbackUrl != nil && UIApplication.shared.canOpenUrl(phoneFallbackUrl!)) {
UIApplication.shared.open(phoneFallbackUrl!, options:[String:Any]()) { (success) in
if(!success) {
// Show an error message: Failed opening the url
}
}
} else {
// Show an error message: Your device can not do phone calls.
}

# Swift #

if let url = NSURL(string: "tel://\(number)"),
UIApplication.sharedApplication().canOpenURL(url) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}

Use openurl.

For making a call in swift 5.1, just use the following code: (I have tested it in Xcode 11)

```
let phone = "1234567890"
if let callUrl = URL(string: "tel://\(phone)"), UIApplication.shared.canOpenURL(callUrl) {
UIApplication.shared.open(callUrl)
}
```
Edit: For Xcode 12.4, swift 5.3, just use the following:

```UIApplication.shared.open(NSURL(string: "tel://555-123-1234")! as URL)```

Make sure that you import UIKit, or it will say that it cannot find UIApplication in scope.
-----------------------

Probably the *mymobileNO.titleLabel.text* value doesn't include the scheme **tel://**

Your code should look like this:

**ObjectiveC**

NSString *phoneNumber = [@"tel://" stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];

**Swift**

if let url = URL(string: "tel://\(mymobileNO.titleLabel.text))") {
UIApplication.shared.open(url)
}