iphone - 我想添加一个删除帖子按钮

Question

前提

我用firebase管理帖子，但是无法实现发帖删除功能。

代码Swift4 postApi

    import Foundation
　　　import FirebaseDatabase

class PostApi {
    var REF_POSTS = Database.database().reference().child("posts")
    var REF_MY_POSTS = Database.database().reference().child("myPosts")

    static var shared: PostApi = PostApi()
    private init() {
    }

    func observePosts(completion: @escaping (PostModel) -> Void) {
        REF_POSTS.observe(.childAdded) { (snapshot) in
            guard let dic = snapshot.value as? [String: Any] else { return }
            let newPost = PostModel(dictionary: dic, key: snapshot.key)
            completion(newPost)
        }
    }

    func observePost(withPostId id: String, completion: @escaping (PostModel) -> Void ) {
        REF_POSTS.child(id).observeSingleEvent(of: .value) { (snapshot) in
            guard let dic = snapshot.value as? [String : Any] else { return }
            let newPost = PostModel(dictionary: dic, key: snapshot.key)
            completion(newPost)
        }
    }

    func observeMyPosts(withUid uid: String, completion: @escaping (String) -> Void) {
        REF_MY_POSTS.child(uid).observe(.childAdded) { (snapshot) in
            let postId = snapshot.key
            completion(postId)
        }
    }

    func fetchCountPosts(withUid uid: String, completion: @escaping(UInt) -> Void) {
        REF_MY_POSTS.child(uid).observe(.value) { (snapshot) in
            let postCount = snapshot.childrenCount
            completion(postCount)
        }
    }

    func observeLike(withPostId id: String, completion: @escaping (Int) -> Void) {
        REF_POSTS.child(id).observe(.childAdded) { (snapshot) in
            guard let value = snapshot.value as? Int else { return }
            completion(value)
        }
    }


            guard let dic = snapshot?.value as? [String: Any] else { return }
            guard let postId = snapshot?.key else { return }
            let updatedPost = PostModel(dictionary: dic, key: postId)
            onSuccess(updatedPost)
        }
    }

    // 投稿を読み込んで好きな順に並べ替える
    func observeTopPost(completion: @escaping(PostModel) -> Void) {
        REF_POSTS.queryOrdered(byChild: "likeCount").observeSingleEvent(of: .value) { (snapshot) in
            let arraySnapshot = (snapshot.children.allObjects as! [DataSnapshot]).reversed()

            arraySnapshot.forEach({ (child) in
                guard let dic = child.value as? [String: Any] else { return }
                let post = PostModel(dictionary: dic, key: snapshot.key)
                completion(post)
            })
        }
    }

代码Swift4 用户接口(interface)

import Foundation
import FirebaseDatabase
import FirebaseAuth

class UserApi {
    // Adresse Datenbank users
    var REF_USERS = Database.database().reference().child("users")

    // Singleton pattern (Einzelstück Muster)
    static var shared: UserApi = UserApi()
    private init() {
    }

    // Aktuell eingeloggte User ID
    var CURRENT_USER_UID: String? {
        if let currentUserUid = Auth.auth().currentUser?.uid {
            return currentUserUid
        }
        return nil
    }

    // Aktuell eingeloggte User
    var CURRENT_USER: User? {
        if let currentUserUid = Auth.auth().currentUser {
            return currentUserUid
        }
        return nil
    }

    // Lade User mit der id
    func observeUser(uid: String, completion: @escaping (UserModel) -> Void) {
        REF_USERS.child(uid).observeSingleEvent(of: .value) { (snapshot) in
            guard let dic = snapshot.value as? [String: Any] else { return }
            let newUser = UserModel(dictionary: dic)
            completion(newUser)
        }
    }

    // Lade user ohne id
    func observeUser(completion: @escaping (UserModel) -> Void ) {
        REF_USERS.observe(.childAdded) { (snapshot) in
            guard let dic = snapshot.value as? [String: Any] else { return }
            let user = UserModel(dictionary: dic)
            completion(user)
        }
    }

    // Lade den aktuellen User
    func observeCurrentUser(completion: @escaping (UserModel) -> Void ) {
        guard let currentUserUid = CURRENT_USER_UID else { return }
        REF_USERS.child(currentUserUid).observeSingleEvent(of: .value) { (snapshot) in
            guard let dic = snapshot.value as? [String: Any] else { return }
            let currentUser = UserModel(dictionary: dic)
            completion(currentUser)
        }
    }

    // User im Suchfeld laden
    func queryUser(withText text: String, completion: @escaping(UserModel) -> Void ) {
        REF_USERS.queryOrdered(byChild: "username_lowercase").queryStarting(atValue: text).queryEnding(atValue: text + "\u{f8ff}").queryLimited(toLast: 5).observeSingleEvent(of: .value) { (snapshot) in
            snapshot.children.forEach({ (data) in
                let child = data as! DataSnapshot
                guard let dic = child.value as? [String: Any] else { return }
                let user = UserModel(dictionary: dic)
                completion(user)
            })
        }
    }
}

补充信息

如果我们有其他缺失信息，我们将添加其他信息。由于我经常在4周内不理解Swift，我希望你用具体的代码等告诉我。

引用资料 How to delete a child from Firebase (Swift)

Answer 1

这是对答案的大胆猜测，因为尚不清楚 OP 如何确定要删除的节点。

让我们假设一个 tableView 显示在 UI 中并且用户滑动了一个 row_to_delete。我们还将假设我们有一组 Post 对象，这些对象具有来自快照的“key”属性。

然后要删除帖子，请从数据源数组中检索帖子并获取它的键。然后为该 ans removeValue 构建一个引用。

let postToDelete = postsArray[row_to_delete]
let postKey = postToDelete.key
let ref = your_firebase.child("posts").child(postKey)
ref.removeValue()