Firebase 数据库：从孩子那里获取所有孩子的数据

Question

I want to be able to differentiate my content between type and get all other data but I don't know how to get past the IDs after content .

I want every data from -NBC22CHEk3ebNxdlYOi, -NBC-Nfz6e5yppFlxU7S, -NBC1TVY-TzT7yfrSA3o and later be possible to get new video id's from existing and new users

/content/{user_uid}/{content_uid}

If I want only the content from one user then I can write this:

val uid = FirebaseAuth.getInstance().uid
// url is just a placeholder for the real database url
val ref = FirebaseDatabase.getInstance(url).getReference("/content/$uid")

Answer 1

I want all content with the type value "video" from all users.

That's actually not possible with your actual database structure. All queries in the Realtime Database work on a flat list of nodes, where the value on which you want to perform the filter/order on, must be at a fixed path under each direct child node.

Unfortunately, that's not your case, because you have a dynamic key under the content node, which is actually the UID, and which also means that you can't filter on type . If you want to allow that, then you have to create a flat list of videos, where the UID (SIH5...89j1) is a property inside each video object. This is already achieved since publisher_uid holds exactly that.

In short, remove the extra (UID) level from your database tree. Or, leave it as it is, and denormalize the data. Basically create a new structure like this:

/content/{content_uid}

See, there is no UID involved. If you're new to NoSQL databases, this might sound some kind of weird, but it's actually a quite common practice.