ios - 单击Swift中的标签

我想做一个标签，点击它可以拨打电话。我知道iOS拥有此选项，但是如何在Swift中做到这一点？

我在ObjC中发现了如何做:

-(IBAction)callPhone:(id)sender {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"tel:2135554321"]];
}

有人可以帮我吗？

最佳答案

UIApplication.sharedApplication().openURL(NSURL(string: "tel://2135554321"))

示例

if let  CallURL:NSURL = NSURL(string:"tel://\(yourMobileNUmber)") {
let application:UIApplication = UIApplication.sharedApplication()
if (application.canOpenURL( CallURL)) {
  application.openURL( CallURL);
  }
else
 {
   // your number not valid
   let tapAlert = UIAlertController(title: "Alert!!!", message: "Your mobile number is invalid", preferredStyle: UIAlertControllerStyle.Alert)
tapAlert.addAction(UIAlertAction(title: "OK", style: .Destructive, handler: nil))
 self.presentViewController(tapAlert, animated: true, completion: nil)
  }
 }

类型2

  // add gesture to your Label
 var tapGesture = UITapGestureRecognizer(target: self, action: Selector("handleTap:"))
 yourLabelName.userInteractionEnabled=true
 yourLabelName.addGestureRecognizer(tapGesture)

// handle the function of UILabel
func handleTap(sender:UITapGestureRecognizer){
     if let  CallURL:NSURL = NSURL(string:"tel://\(yourMobileNUmber)") {
let application:UIApplication = UIApplication.sharedApplication()
if (application.canOpenURL( CallURL)) {
  application.openURL( CallURL);
  }
 else
 {
   // your number not valid
   let tapAlert = UIAlertController(title: "Alert!!!", message: "Your mobile number is invalid", preferredStyle: UIAlertControllerStyle.Alert)
tapAlert.addAction(UIAlertAction(title: "OK", style: .Destructive, handler: nil))
 self.presentViewController(tapAlert, animated: true, completion: nil)
  }
 }
}

关于ios - 单击Swift中的标签，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/31375688/