tags:

views:

117

answers:

3

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?

A: 

no it's not possible

elmac
ok thx but there is a application where it is happening i m giving you reference for that appshttp://www.codegoo.com/page/baby-monitorhere it is going on???
tech.samar
Serious? 0% accept rate?
Chuck Conway
A: 

Try this:

UIWebView *callingWebview;

[callingWebview loadRequest:[NSURLRequest requestWithURL:]];

hawshy
A: 

Hawsky when I try with this,A alert is dispalyed to trigger call or not ,requesting user.I want to trigger call with out the user input.How to do...

Dinesh Kumar