So few day ago I moved my apps posts to cloud firestore from realtime database because of the more flexable querying that firestore apparently has. I have been modifying my code to work with firestore as the new database. Now I ran into a problem. How can I retrieve every document within "Posts" collection that has "Likes" collection that contains a specifically named document? I don't know if that makes sense so here's an image showcasing the problem:
I had a working query for firebase realtime database but can't figure out how to do it with firestore. Is it possible or do I have to change the structure of my database? If I do have to change the structure, what would be the most convenient way of getting the same result?
This is currently not possible with collection group queries. You might be tempted to put a filter on the query for FieldPath.documentId() == 'your-doucment-id', but that just isn't supported. FieldPath.documentId() is actually a token that refers to to fully unique document id, which includes the entire path of the document.
The workaround is to store the document ID as a field in the document, and filter on the contents of that field instead. So, if the document had a field called id with the ID of the document, you could query for collectionGroup("Likes").whereEquals('id', 'your-document-id').
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How does one filter a Firestore collection by a document reference in the web console? For example, I have a collection that references another collection document called challenges.
But no matter what combination I use to try to filter that collection by the challenge ID it returns no results. I need to be able to do this so I can look at all of the documents in this collection that reference back to that particular challenge reference.
I am assuming this is possible in the console, but maybe it is not?
I use this code to get a collection snapshot from Firestore.
firestore().collection('project').where('userID', '==', authStore.uid).onSnapshot(onResult, onError);
This returns a huge amount of data, but I only need a few fields. Is it possible to query only a specific field? For example, if I only need the projectName and the creationDate fields.
Is it possible to query only a specific field?
No, that is not possbile. A Firestore listener fires on the document level. This means that you'll always get the entire document.
For example if I only need the projectName and the creationDate fields.
You cannot only get the value of a specific set of fields. It's the entire document or nothing. If you, however, only need to read those values and nothing more, then you should consider storing them in a separate document. This practice is called denormalization, and it's a common practice when it comes to NoSQL databases.
You might also take into consideration using the Firebase Realtime Database, for the duplicated data.
My cloud firestore database has an "orders" collection and in HTML I have a 'save' button to add document(s) into that "orders" collection upon clicking. Now, using add will assign auto-generated ID for each document.
What if I want to customise such ID by timestamp? So that the document created yesterday will be assigned an index as '1', and the following document created will be '2', etc...
What you're trying to do is not compatible with the way Cloud Firestore was designed. Firestore will not assign monotonically increasing numbers for document IDs. This just doesn't scale massively as required by Firestore and would introduce performance bottlenecks.
If you want to be able to sort documents by timestamp, the best strategy is to add a timestamp field to each document, then use that field in an ordered query.
Note that you could try to write a lot of code to get this done the way you want, but you are MUCH better off accepting the random IDs and using fields to filter and order data.
in some case, when you need to save several docs in different collection due to an event occurs, it's better to same all docs with same id in different collections with single firestore server's timestamp. you get the timestamp like below:
const admin = require('firebase-admin')
const ts = admin.firestore.Timestamp.now().toMillis().toString()
by doing this, when you need to read all those docs, you only need to query once to get timestamp, then read all other doc by timestamp directly.
it should be faster than query the timestamp inside document fields for each collections
I'd like my web app router slugs to correspond to my Firestore documents data.
For example:
www.mysite.com/restaurants/burger-king
/restaurants <- Firestore Collection
/restaurants/mcdonalds <- Firestore Document
/restaurants/burger-king <- Firestore Document
This is easy enough, as I can assign the name as a slug-friendly UID in Firestore. The difficulty arises with CRUD functionality. I need to be able to rename my item titles, but Firestore does not permit you to rename indexes, which is the issue I'm facing.
One SO solution I saw was to delete the old record and creates a new one at the updated index. That's problematic for me, because sub-collections would be hard to transfer from the client side.
Are there more elegant solutions?
You don't have to identify a document by its ID. If you're able to ensure uniqueness of a document field value, you could instead query a collection for an ID value in a known field, then use the results of that query to satisfy your REST API. Then, you can change the value of that document field as often as you want, in order to satisfy required changes to the public API.
I am working on angular 5 with firebase firestore.I inserted data into fire-store collection/document/collection/document/collection/document then set data the last document.But at the time retrieving data I am giving only first collection name then I want complete root data.but I am getting only field in it.
https://firebase.google.com/docs/firestore/query-data/get-data
checkout:List subcollections of a document
As we know querying in Cloud Firestore is shallow by default. This type of query isn't supported, although it is something Google may consider in the future.