本文介绍了以编程方式拨打带有访问代码的电话号码的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

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:


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];
}

这篇关于以编程方式拨打带有访问代码的电话号码的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

09-09 06:07