CODEWITHSUNDEEP
iosiphoneobjective-ccocoa-touch
Dharmendra Vaishnav
Dharmendra Vaishnav

Reputation: 45

open dialer in ios on click of button

I tried following code but it is not working 

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel://%@",@"55698"]]];

Please provide help which support iOS7 SDK.

Upvotes: 4

Views: 10646

Answers (7)

Nikolay DS
Nikolay DS

Reputation: 1477

Updated for Swift and iOS 10 and up, where phone number is a string

        //call Phone dialer
        if  let url = URL(string: "tel://\(phoneNumber)"), UIApplication.shared.canOpenURL(url)  {
            UIApplication.shared.open(url, options: [:], completionHandler: nil)
        }

Upvotes: 1

Vikas Rajpurohit
Vikas Rajpurohit

Reputation: 123

In Swift 2.0 - >= iOS 7.0 , you can use the following:

    let url:NSURL = NSURL(string: "tel://9809088798")!

    if (UIApplication.sharedApplication().canOpenURL(url))
    {
    UIApplication.sharedApplication().openURL(url)
    }

Upvotes: 3

Manab Kumar Mal
Manab Kumar Mal

Reputation: 21378

Above answers are fine. Still please log, you have some Blank space(@" ") left in front or back of the Phone number string. I was facing same problem and getting the error like this:

LaunchServices: ERROR: There is no registered handler for URL scheme (null)

Then I have added that code and Open the Prompt dialler like the Below code

NSString *strPhone=[[[dictEvent valueForKey:@"app_contact_value"] stringByReplacingOccurrencesOfString:@"+" withString:@""] stringByReplacingOccurrencesOfString:@" " withString:@""];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"telprompt://%@",strPhone]]];

and now it is Running fine.

You can also directly call with phone number like other answers:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"telprompt://%@",strPhone]]];

Upvotes: 4

Rajesh Loganathan
Rajesh Loganathan

Reputation: 11217

Try this simple code:

 NSURL *URL = [NSURL URLWithString:@"tel://55698"];
 [[UIApplication sharedApplication] openURL:URL];

Try it out in device not in simulator...

Upvotes: 3

Himanshu Joshi
Himanshu Joshi

Reputation: 3399

Do not use tel:// 

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:55698"]]];

Upvotes: 1

Preson
Preson

Reputation: 244

You can place a call through the phone app using 

NSString *phoneNumber = @"tel://472490168092";
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];

You cannot open the keypad within the phone app in iOS.You can design your own phone pad in your view controller, add tones for keys. That's the only way to achieve this AFAIK.

Upvotes: 2

DharaParekh
DharaParekh

Reputation: 1730

Remove // after tel: in your code

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"tel:55698"]];

OR

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:%@",@"55698"]]];

Upvotes: 7

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