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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?

Bill the Lizard
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tech.samar
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2 Answers2

3

Try this:

UIWebView *callingWebview;

[callingWebview loadRequest:[NSURLRequest requestWithURL:]];

hawshy
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no it's not possible

elmac
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  • ok thx but there is a application where it is happening i m giving you reference for that apps http://www.codegoo.com/page/baby-monitor here it is going on??? – tech.samar Aug 02 '10 at 11:23
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    Yeah it is possible.Through this code.. [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"telprompt:XXXXXX"]]; – tech.samar May 11 '11 at 11:04