CODEWITHSUNDEEP
iphoneobjective-ciosios4
Ranjit
Ranjit

Reputation: 4646

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

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...

Upvotes: 8

Views: 10391

Answers (6)

Mani
Mani

Reputation: 305

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]];

Upvotes: 0

Anoop Singh
Anoop Singh

Reputation: 43

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

Upvotes: 0

Sambit K.
Sambit K.

Reputation: 55

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.

Upvotes: 1

Sandhya Shettigar
Sandhya Shettigar

Reputation: 299

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];

Upvotes: -1

Sandhya Shettigar
Sandhya Shettigar

Reputation: 299

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.

Upvotes: 18

Lokus001
Lokus001

Reputation: 159

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

Upvotes: -4

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