如何在iOS 9中打开手机拨号器？

Question

我找到了需要添加一些代码的解决方案info.plist.我这样做如下:

<key>LSApplicationQueriesSchemes</key>
<array>
    <string>tel</string>
</array>

仍然没有帮助.我收到此错误:

"-canOpenURL:URL失败:"tel:// 4806501708" - 错误:"此应用程序不允许查询方案电话"

我打开拨号器的代码:

NSString *phoneNumber = [@"tel://" stringByAppendingString:lblVenPhoneValue.text];
if ([UIApplication.sharedApplication canOpenURL:[NSURL URLWithString:phoneNumber]]) {
        [UIApplication.sharedApplication openURL:[NSURL URLWithString:phoneNumber]];

我需要做什么？
提前致谢

Answer 1

你在设备上测试这个吗？,因为这不适用于模拟器.设备也应该有SIM卡.

确认以上后尝试以下

在info.plist中

<key>LSApplicationQueriesSchemes</key>
<array>
<string>tel</string>
<string>telprompt</string>
</array>

哪里想打开手机拨号器

 [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:%@",@"digits"]]]; 

要么

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"telprompt:%@",@"digits"]]];

Answer 2

只需从@“tel://”中删除“//”就可以了

NSString *phoneNumber = [@"tel:" stringByAppendingString:lblVenPhoneValue.text];
if ([UIApplication.sharedApplication canOpenURL:[NSURL URLWithString:phoneNumber]]) {
        [UIApplication.sharedApplication openURL:[NSURL URLWithString:phoneNumber]];

为了进行更多更好的检查，您可以使用

if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:[NSString stringWithFormat:phoneNumber]]])
{
  CTTelephonyNetworkInfo *networkInfo = [CTTelephonyNetworkInfo new];
  CTCarrier *carrier = [networkInfo subscriberCellularProvider];
  NSString *_code = [carrier mobileNetworkCode];
  if(_code)
  {
    [[UIApplication sharedApplication] openURL:phoneNumber]];
  }
  else
  {
    UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"no_sim" message:@"" delegate:nil cancelButtonTitle:@"ok" otherButtonTitles:nil, nil];
    [alert show];
  }
}
else
{
  UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"alert" message:@"alert_device_not_support" delegate:nil cancelButtonTitle:@"ok" otherButtonTitles:nil, nil];
  [alert show];
}