从 Firebase 快速加载图像

Question

有人知道为什么这不起作用吗？即使数据库和存储中有项目，tableView 也是空的并且不显示任何内容。在我实现从存储中加载图像之前，这很好用) 但是，如果我现在将其取出，它将无法访问recipeImage，因为它是在闭包中声明的。它不工作是因为它在关闭中吗？如果是这样，我该如何解决？

let parentRef = Database.database().reference().child("Recipes")
    let storage = Storage.storage()

    parentRef.observe(.value, with: { snapshot in

        //Processes values received from server
        if ( snapshot.value is NSNull ) {

            // DATA WAS NOT FOUND
            print("– – – Data was not found – – –")

        } else {

            //Clears array so that it does not load duplicates
            food = []

            // DATA WAS FOUND
            for user_child in (snapshot.children) {

                let user_snap = user_child as! DataSnapshot
                let dict = user_snap.value as! [String: String?]

                //Defines variables for labels
                let recipeName = dict["Name"] as? String
                let recipeDescription = dict["Description"] as? String
                let downloadURL = dict["Image"] as? String

                let storageRef = storage.reference(forURL: downloadURL!)

                storageRef.getData(maxSize: 1 * 1024 * 1024) { (data, error) -> Void in

                    let recipeImage = UIImage(data: data!)

                    food.append(Element(name: recipeName!, description: recipeDescription!, image: recipeImage!))
                }
            }
            self.tableView.reloadData()
        }
    })

Answer 1

移动

self.tableView.reloadData()

之后

food.append(Element(name: recipeName!, description: recipeDescription!, image: recipeImage!))