在 iOS 中以编程方式拨打带有访问代码的电话号码

Dial a phone number with an access code programmatically in iOS(在 iOS 中以编程方式拨打带有访问代码的电话号码)

问题描述

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?

How can I dial a phone number that includes a number and access code programmatically in iOS?

例如:

号码:900-3440-567
访问代码:65445

number: 900-3440-567
Access Code: 65445

推荐答案

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];
}

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