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 allo         


        

Answer 1

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]];

Answer 2

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];

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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