从iphone应用程序以编程方式拨打电话,并在结束通话后返回应用程序 [英] Making a call programatically from iphone app and returning back to the app after ending the call
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问题描述
我正试图通过我的iphone应用程序发起呼叫,我按照以下方式进行了调用..
I am trying to initiate a call from my iphone app,and I did it the follwing 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];
}
}
上面的代码.. 我能够成功拨打电话..但是当我结束通话时,我无法返回我的应用
所以,我想要知道如何实现,那也请告诉我如何使用webview发起呼叫...
So,I want to know how to achieve,that..also please tell me how we can initiate a call using webview...
谢谢和问候
Ranjit
Thanks&Regards Ranjit
推荐答案
这是我的代码:
NSURL *url = [NSURL URLWithString:@"telprompt://123-4567-890"];
[[UIApplication sharedApplication] openURL:url];
使用此选项,以便在通话结束后它将返回app。
Use this so that after call end it will return to app.
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