Ios 快速从firebase存储中删除图像

我有一个功能，可以通过滑动tableView单元格从firebase数据库中删除对象，但是，我的tableView单元格也包含保存在firebase存储中的图像，我希望在从数据库中删除数据时也从存储中删除图像，我如何才能做到这一点

删除代码：

func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCellEditingStyle, forRowAt indexPath: IndexPath) {
    if editingStyle == .delete {

        let name = food[indexPath.row].name
        let ref = Database.database().reference().child("Recipes")

        ref.queryOrdered(byChild: "Name").queryEqual(toValue: name).observe(.childAdded, with: { (snapshot) in
            //Removes deleted cell from firebase
            snapshot.ref.removeValue(completionBlock: { (error, reference) in
                if error != nil {
                    print("There has been an error: \(error)")
                }
                //Removes deleted cell from array
                food.remove(at: indexPath.row)
                //Removes deleted cell from tableView
                tableView.deleteRows(at: [indexPath], with: .left)
            })
        })
    }
}

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

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

        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()
                }
            }
        }
    })

加载代码：

func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCellEditingStyle, forRowAt indexPath: IndexPath) {
    if editingStyle == .delete {

        let name = food[indexPath.row].name
        let ref = Database.database().reference().child("Recipes")

        ref.queryOrdered(byChild: "Name").queryEqual(toValue: name).observe(.childAdded, with: { (snapshot) in
            //Removes deleted cell from firebase
            snapshot.ref.removeValue(completionBlock: { (error, reference) in
                if error != nil {
                    print("There has been an error: \(error)")
                }
                //Removes deleted cell from array
                food.remove(at: indexPath.row)
                //Removes deleted cell from tableView
                tableView.deleteRows(at: [indexPath], with: .left)
            })
        })
    }
}

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

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

        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()
                }
            }
        }
    })

如果有人能帮我解决我问的关于同一个应用程序的另一个问题，我也会非常感激：

编辑：

func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCellEditingStyle, forRowAt indexPath: IndexPath) {
    if editingStyle == .delete {

        let name = food[indexPath.row].name
        let ref = Database.database().reference().child("Recipes")

        ref.queryOrdered(byChild: "Name").queryEqual(toValue: name).observe(.childAdded, with: { (snapshot) in
            //Removes deleted cell from firebase
            snapshot.ref.removeValue(completionBlock: { (error, reference) in
                if error != nil {
                    print("There has been an error: \(error)")
                }
                //Removes deleted cell from array
                food.remove(at: indexPath.row)
                //Removes deleted cell from tableView
                tableView.deleteRows(at: [indexPath], with: .left)
            })
        })
    }
}

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

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

        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()
                }
            }
        }
    })

从firebase加载对象时，我已将URL添加到数组中：

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

这就是我试图用来删除的内容：

let storage = Storage.storage()
let storageRef = storage.reference()
let desertRef = storageRef.child(food[indexPath.row].downloadURL)

//Removes image from storage
desertRef.delete { error in
    if let error = error {
        print(error)
    } else {
        // File deleted successfully
    }
}

我不认为它能找到图像，但是。。。我得到这个错误：

Error Domain=FIRStorageErrorDomain Code=-13010“Object https:/firebasestorage.googleapis.com/v0/b/recipe-app-1b76e.appspot.com/o/B74F604B-68FD-45BB-ABDB-150B03E83A2A.png？alt=media&token=ae2643c4-6479-4dc8-b389-d04caac98392不存在。”用户信息={object=https:/firebasestorage.googleapis.com/v0/b/recipe-app-1b76e.appspot.com/o/B74F604B-68FD-45BB-ABDB-150B03E83A2A.png？alt=media&token=ae2643c4-6479-4dc8-b389-d04caac98392，bucket=recipe-app-1b76e.appspot.com，ResponseBody={ “错误”：{ “代码”：404， “消息”：“未找到。无法删除对象” } },

---

基于图像位置创建参考

// Create a reference to the file to delete
let imageRef = storageRef.child("image.png")

// Delete the file
imageRef.delete { error in
  if let error = error {
    // Uh-oh, an error occurred!
  } else {
    // File deleted successfully
  }
}

在下一行之前添加上面的代码

food.remove（位于：indexPath.row）

---

！

解决了！以下是对我有效的方法：

let storage = Storage.storage()
let url = food[indexPath.row].downloadURL
let storageRef = storage.reference(forURL: url)

//Removes image from storage
storageRef.delete { error in
    if let error = error {
        print(error)
    } else {
        // File deleted successfully
    }
}

---

也许解决方法是，您的食物对象添加下载url并删除storageRef.let-desertRef=storageRef.child（food.downloadUrl）desertRef.delete…@EmreYILMAZ嘿，请查看我的question@EmreYILMAZ没关系，我修正了！请检查我的其他问题！嘿，请看我问题底部的编辑没关系，我修正了！请检查我的其他问题！