ios - 在 iOS 中以编程方式调用带有访问代码的电话号码

Question

如何在 iOS 中以编程方式调用包含号码和访问代码的电话号码？

例如:

number: 900-3440-567
Access Code: 65445

Answer 1

UIDevice *device = [UIDevice currentDevice];
if ([[device model] isEqualToString:@"iPhone"] ) {
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:130-032-2837"]]];
} else {
    UIAlertView *notPermitted=[[UIAlertView alloc] initWithTitle:@"Alert" message:@"Your device doesn't support this feature." delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
    [notPermitted show];
    [notPermitted release];
}