使用 Query 读取 Firebase 数据库返回 null

Question

在这里，我尝试通过匹配我从事件节点获得的 userId 和 eventId 来获取喜欢的事件节点密钥。我做了一些代码，但查询在 likeEventKey 上返回 null。

这是我的事件节点 我试图在 onDataChange 方法中包含循环和不循环。结果是一样的。它给我null。

这里的节点是

the node

    private String likedEventKey;

 /*
     * Attach the data to the UI
     * */
    private void showEventData(Event event) {
        textView.setText(event.getEventKey() + ", " + currentUser.getUid() + "," + likedEventKey);
    }

/*
     * Method for checking like event status in Db based on EventKey
     * */
    private void checkLikeEventStatusInDb() {
        Query query = mLikedEventDbReference
                .child("liked-event")
                .orderByChild("eventId")
                .equalTo(event.getEventKey());
        query.addListenerForSingleValueEvent(new ValueEventListener() {
            @Override
            public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
                for (DataSnapshot ds : dataSnapshot.getChildren()) {
                    likedEvent = ds.getValue(LikedEvent.class);
                    if (likedEvent != null) {
                        String eventId = likedEvent.getEventId();
                        String eventKey = event.getEventKey();
                        if (eventId.equalsIgnoreCase(eventKey)) {
                            checkUsernameInLikeEventStatusInDb();
                        }
                    }
                }
            }

            @Override
            public void onCancelled(@NonNull DatabaseError databaseError) {
                Timber.e("Error while fetch letch liked event %s",
                        databaseError.getMessage());
            }
        });
    }

    /*
     * Method for checking like event status in Db based on UserId
     * */
    private void checkUsernameInLikeEventStatusInDb() {
        Query query = mLikedEventDbReference
                .child("liked-event")
                .orderByChild("userId")
                .equalTo(currentUser.getUid());
        query.addListenerForSingleValueEvent(new ValueEventListener() {
            @Override
            public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
                for (DataSnapshot ds : dataSnapshot.getChildren()) {
                    likedEvent = ds.getValue(LikedEvent.class);
                    if (likedEvent != null) {
                        likedEvent.setLikedEventKey(ds.getKey());
                        String userIdInLikedEvent = likedEvent.getUserId();
                        if (userIdInLikedEvent.equalsIgnoreCase(currentUser.getUid())) {
                            likedEventKey = likedEvent.getLikedEventKey();
                            Timber.d("likedEventKey: %s", likedEventKey);
                        }
                    }
                }
            }

            @Override
            public void onCancelled(@NonNull DatabaseError databaseError) {
                Timber.e("Error while fetch letch liked event %s",
                        databaseError.getMessage());
            }
        });
    }

我的预期回报是我得到了基于 userId 和 eventId 的喜欢事件子键。

Answer 1

下面的查询只会返回一个 liked-event 节点（检查here）。因此onDataChange() 中的 for 循环没有任何意义。它在 liked-event

内循环遍历子项

Query query = mLikedEventDbReference
     .child("liked-event")
     .orderByChild("userId")
     .equalTo(currentUser.getUid());

通过如下所述更改代码，我想您将能够实现您所需要的。

    Query query = mLikedEventDbReference
            .child("liked-event")
            .orderByChild("eventId");

    query.addListenerForSingleValueEvent(new ValueEventListener() {
        @Override
        public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
            for (DataSnapshot ds : dataSnapshot.getChildren()) {
                likedEvent = ds.getValue(LikedEvent.class);
                if (likedEvent != null) {
                    String eventId = likedEvent.getEventId();
                    String eventKey = ds.getKey();
                    if (eventId.equalsIgnoreCase(eventKey)) {
                        checkUsernameInLikeEventStatusInDb();
                    }
                }
            }
        }
    });

你也可以用类似的方法改变其他方法。