如何从ios应用程序拨打电话号码？

Question

我试图通过以下方式从ios应用程序拨打电话号码:虽然方法被调用,但它无法正常工作:

-(IBAction)callPhone:(id)sender {

        NSString *phoneCallNum = [NSString stringWithFormat:@"tel://%@",listingPhoneNumber ];

        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneCallNum]];

        NSLog(@"phone btn touch %@", phoneCallNum);
    }

NSLog 输出:手机btn touch tel:// + 39 0668806972

Answer 1

你的代码是正确的.你检查过真正的设备吗？此功能在模拟器中不起作用.

试试这个,

NSString *phNo = @"+919876543210";
NSURL *phoneUrl = [NSURL URLWithString:[NSString  stringWithFormat:@"telprompt:%@",phNo]];

    if ([[UIApplication sharedApplication] canOpenURL:phoneUrl]) {
        [[UIApplication sharedApplication] openURL:phoneUrl];
    } else
    {
        calert = [[UIAlertView alloc]initWithTitle:@"Alert" message:@"Call facility is not available!!!" delegate:nil cancelButtonTitle:@"ok" otherButtonTitles:nil, nil];
        [calert show];
    }

**Swift 3版**

if let url = URL(string: "telprompt:\(phoneNumber)") {
  if UIApplication.shared.canOpenURL(url) {
    UIApplication.shared.openURL(url)
  }
}

Answer 2

电话无法在模拟器/ iPod/iPad上运行,您需要在具有活动SIM卡的iPhone上运行该应用程序.

还有用于调用电话应用程序的URL方案tel:<phone_number>.请参阅Apple文档.

理想情况下,您应检查设备是否具有电话模块,然后执行openURL:呼叫.使用此代码执行检查,

if([[UIApplication sharedApplication] canOpenURL:callUrl]) {
    [[UIApplication sharedApplication] openURL:callUrl];
}
else {
    //Show error message to user, etc.
}

Answer 3

使用以下方法拨打电话:

 NSString *phoneNumber = [@"tel://" stringByAppendingString:number];
 [[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];