在这里,我尝试通过匹配从事件节点获取的 userId 和 eventId 来获取喜欢的事件节点 key 。我做了一些代码,但查询在 likedEventKey 上返回 null。
这是我的事件节点 我试图在 onDataChange 方法中包含 loopin 和不包含 loopin。结果是一样的。它给我空。
这里的节点是
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());
}
});
}
我的预期返回是我根据 userId 和 eventId 得到了喜欢的事件子键。
最佳答案
下面的查询只会返回一个喜欢的事件节点(检查here)。因此 onDataChange()
中的 for 循环没有任何意义。它在喜欢的事件
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();
}
}
}
}
});
您也可以像这样更改其他方法。
关于java - 使用 Query 读取 Firebase 数据库返回 null,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57217315/