在 iOS 中以编程方式拨打带有访问代码的电话号码

如何在 iOS 中以编程方式拨打包含号码和访问代码的电话号码？

例如：

号码：900-3440-567
访问代码：65445

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];
}