What’s the correct URL for placing a call on an iPhone?

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闹比i
闹比i 2020-12-11 06:55

Is the code for calling on the iPhone automatically

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@\"tel:11111111111\"]]);
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5条回答
  • 2020-12-11 07:10

    From trying on an iPhone, tel://123456789 is the way to go. The tel:123456789 option is not even recognized, at least by the Safari URL bar.

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  • 2020-12-11 07:14

    Your second line is fine and will work.

    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"tel://1111111111"]]);
    
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  • 2020-12-11 07:17

    you can only call from Iphone device not from ipad/ipod, and you can dial number from iphone like bellow code:-

    NSString *value=@"your number";
    NSURL *url = [[ NSURL alloc ] initWithString:[NSString stringWithFormat:@"tel://%@",value]];
    [[UIApplication sharedApplication] openURL:url]; 
    
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  • 2020-12-11 07:28

    SwiftArchitect's answer doesn't fit all. I wanted to actually initiate an automatic call, not prompt.

    So there is a difference between tel and telprompt.

    tel: actually initiates the call.

    if let url = URL(string: "tel:\(phoneNumber)") {
      if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.openURL(url)
      }
    }
    

    telprompt: prompts for call or cancel.

    if let url = URL(string: "telprompt:\(phoneNumber)") {
      if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.openURL(url)
      }
    }
    

    I didn't know the difference. Question also asks for calling. So this kind of answer would have helped me save time.

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  • 2020-12-11 07:32

    Too many answers with conflicting comments.

    (slash, no slash, semicolumn, tel, telprompt ?)

    Swift, one size fits all:

    if let phoneURL = NSURL(string: "telprompt:\(phoneNumber)") {
        if UIApplication.sharedApplication().canOpenURL(phoneURL) {
            UIApplication.sharedApplication().openURL(phoneURL)
        }
    }
    
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