Making a call programmatically from iPhone app and returning back to the app after ending the call

Question

I am trying to initiate a call from my iphone app,and I did it the following way..

-(IBAction) call:(id)sender
{

    UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Call Besito" message:@"\n\n\n"
                                                   delegate:self cancelButtonTitle:@"Cancel"  otherButtonTitles:@"Submit", nil];


    [alert show];
    [alert release];
}

- (void)alertView:(UIAlertView *)alertView willDismissWithButtonIndex:(NSInteger)buttonIndex
{
    if (buttonIndex != [alertView cancelButtonIndex])
    {
      NSString *phone_number = @"0911234567"; // assing dynamically from your code
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString    stringWithFormat:@"tel:%@", phone_number]]]; 
         NSString *phone_number = @"09008934848";
        NSString *phoneStr = [[NSString alloc] initWithFormat:@"tel:%@",phone_number];
        NSURL *phoneURL = [[NSURL alloc] initWithString:phoneStr];
        [[UIApplication sharedApplication] openURL:phoneURL];
        [phoneURL release];
        [phoneStr release];
    }
}

by the above code..I am able to successfully make a call..but when I end a call,I'm not able to return to my app

So, I want to know how to achieve,that..also please tell me how we can initiate a call using webview...

This question answers your question in detail. Simply use a uiwebview to place call instead of openURL: http://stackoverflow.com/questions/5317783/return-to-app-behavior-after-phone-call-different-in-native-code-than-uiwebview — Bushra Shahid, Sep 26 '11 at 08:37

score 18 · Accepted Answer · answered Jul 18 '12 at 06:29

18

This is my code :

NSURL *url = [NSURL URLWithString:@"telprompt://123-4567-890"]; 
[[UIApplication  sharedApplication] openURL:url];

Use this so that after call end it will return to app.

answered Jul 18 '12 at 06:29

Sandhya Shettigar

299
4
12

welcom, its telprompt string that should write so that we can get back to our app after call ends. – Sandhya Shettigar Jul 18 '12 at 06:41
can I know what all you have implemented? – Ranjit Jul 18 '12 at 07:02
what all means if you specifically ask I can tell whether I did or not. – Sandhya Shettigar Jul 18 '12 at 07:05
Ok, thats fine, actually I am also implementing the drawing feature, I have my code working, but their is a little flaw during erasing, can you please look at this link http://stackoverflow.com/questions/11502320/eraser-not-working-in-ios-drawing – Ranjit Jul 18 '12 at 07:11
is there a way to capture callbacks if call fails due to low balance or network signals? – Rajan Maheshwari Feb 24 '15 at 10:42

score 1 · Answer 2 · edited Sep 06 '13 at 10:48

1

UIWebView *callWebview = [[UIWebView alloc] init];
NSURL *telURL = [NSURL URLWithString:@"tel:+9196815*****"];
[callWebview loadRequest:[NSURLRequest requestWithURL:telURL]];

Please try this it will definitely let you Back to Your App.

edited Sep 06 '13 at 10:48

Soner Gönül

97,193
102
206
364

answered Sep 06 '13 at 10:27

Sambit K.

55
6

score 0 · Answer 3 · edited Feb 27 '15 at 06:19

following code will not return to your app [no alertview will show before make call]

UIWebView *callWebview = [[UIWebView alloc] init];
NSURL *telURL = [NSURL URLWithString:[NSString stringWithFormat:@"tel://%@",phoneNumber]];
[callWebview loadRequest:[NSURLRequest requestWithURL:telURL]];

following code will return to your app "alertview will show before make call"

UIWebView *callWebview = [[UIWebView alloc] init];
NSURL *telURL = [NSURL URLWithString:[NSString stringWithFormat:@"telprompt://%@", phoneNumber]];
[callWebview loadRequest:[NSURLRequest requestWithURL:telURL]];

score 0 · Answer 4 · answered Dec 30 '14 at 07:42

0

NSString *phoneNumber =  // dynamically assigned
NSString *phoneURLString = [NSString stringWithFormat:@"tel:%@", phoneNumber];
NSURL *phoneURL = [NSURL URLWithString:phoneURLString];
[[UIApplication sharedApplication] openURL:phoneURL];

answered Dec 30 '14 at 07:42

Anoop Singh

43
7

score -1 · Answer 5 · answered Jul 18 '12 at 06:59

You can also use webview, this is my code :

NSURL *url = [NSURL URLWithString:@"tel://123-4567-890"];
UIButton *btn = [[UIButton alloc]initWithFrame:CGRectMake(50, 50, 150, 100)];
[self.view addSubview:btn];
UIWebView *webview = [[UIWebView alloc] initWithFrame:CGRectMake(50, 50, 150, 100)];
webview.alpha = 0.0;        
[webview loadRequest:[NSURLRequest requestWithURL:url]];
// Assume we are in a view controller and have access to self.view
[self.view insertSubview:webview belowSubview:btn];
[webview release];
[btn release];

score -4 · Answer 6 · answered Aug 18 '11 at 09:11

-4

It's not possible to return back to the app after ending a call because it's an app.

answered Aug 18 '11 at 09:11

Lokus001

159
1
5

hi lokus,but many say its possible through by using webview..to make a call – Ranjit Aug 18 '11 at 09:18
Are you talking about this post http://stackoverflow.com/questions/3145418/placing-a-call-programmatically-on-iphone-and-return-to-the-same-app-after-hangup ? – Lokus001 Aug 18 '11 at 09:23
ya,,,lokus and laso this http://stackoverflow.com/questions/5317783/return-to-app-behavior-after-phone-call-different-in-native-code-than-uiwebview – Ranjit Aug 18 '11 at 09:25
you can return to application using `telprompt` instead of `tel` – atastrophic Jul 18 '12 at 05:29

Making a call programmatically from iPhone app and returning back to the app after ending the call

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