Reputation: 152
I can make a call from my app by use this APIs.
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"tel:XXXXXX"]];
I would like to return to my app where I left after the users ends the call. Is that possible?
Upvotes: 1
Views: 810
Reputation: 183
Try this:
UIWebView *callingWebview;
[callingWebview loadRequest:[NSURLRequest requestWithURL:]];
Upvotes: 3