oracle

Question

我试图从以下输出：

select count(r.id) and count(r.key) from rules r where lower(r.name) in ('xss attack','dos attack') ;

但是收到错误

  

ORA-00923：未找到FROM关键字

Answer 1

这样的东西？ ：

@IBAction func saveButton(_ sender: Any)
{
    addRestaurant()
}

func addRestaurant()
{
    ref = FIRDatabase.database().reference().child("restaurants")

    let key = ref?.childByAutoId().key

    let name = addName.text

    let phone = addPhone.text

    ref?.child(key!).setValue(["name": name, "phone": phone])

}

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any])
{
    let image = info[UIImagePickerControllerOriginalImage] as? UIImage

    addImage.image = image

    var data = Data()
    data = UIImagePNGRepresentation(image!)!

    let uniqueName = NSUUID().uuidString
    let imageRef = FIRStorage.storage().reference().child("restaurantImage").child("\(uniqueName)")
    imageRef.put(data, metadata: nil).observe(.success){(snapshot) in
        let downloadURL = snapshot.metadata?.downloadURL()?.absoluteString
        let imageDBRef = FIRDatabase.database().reference().child("restaurants/image")
        let key = imageDBRef.childByAutoId().key
        imageDBRef.child(key).setValue(downloadURL)
    }
    self.dismiss(animated: true, completion: nil)
}

Answer 2

我尝试了下面的解决方案：

select (select count(r.id) from rules r where lower(r.name) in ('xss attack','dos attack')) AS name,(select count(r.key) from rules r where lower(r.name) in ('xss attack','dos attack')) AS name from rules;

如果上述查询需要改进，请纠正我。

输出I＆＃39;得到就像

NAME   NAME

 2      2
 2      2

虽然我希望我的输出应该像

一样

NAME        NAME
 2            2