I have a collection of schedules with documents with fields like this
I want to get the route with this highest traffic by running a query that orders by route field and do a count of the same field. My question is how can I count after ordering by a field? This is my query right now
scheduleByRoutes() {
return this.afs.collection('schedules', ref => ref.orderBy('route', 'asc')).snapshotChanges();
}
There is no direct way to count the number of documents of each route "category" returned by your orderBy() query.
You should either:
1/ Count them from your client, iterating on the query results.
or
2/ If you know the different routes upfront, issue a query for each route and use the size() method of each QuerySnapshot. You may use Promise.all() to make these calls in parallel.
or
3/ Maintain some counters for each route in, for example, another collection. For that you would use a set of Cloud Functions that would update the counters upon Creation/Modification/Deletion.
Be aware that approaches #1 and #2 will cost a document read for each document of the collection. If your collection contains a lot of documents, you may use approach #3.
Related
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.
I have a collection where the documents are uniquely identified by a date, and I want to get the n most recent documents. My first thought was to use the date as a document ID, and then my query would sort by ID in descending order. Something like .orderBy(FieldPath.documentId, descending: true).limit(n). This does not work, because it requires an index, which can't be created because __name__ only indexes are not supported.
My next attempt was to use .limitToLast(n) with the default sort, which is documented here.
By default, Cloud Firestore retrieves all documents that satisfy the query in ascending order by document ID
According to that snippet from the docs, .limitToLast(n) should work. However, because I didn't specify a sort, it says I can't limit the results. To fix this, I tried .orderBy(FieldPath.documentId).limitToLast(n), which should be equivalent. This, for some reason, gives me an error saying I need an index. I can't create it for the same reason I couldn't create the previous one, but I don't think I should need to because they must already have an index like that in order to implement the default ordering.
Should I just give up and copy the document ID into the document as a field, so I can sort that way? I know it should be easy from an algorithms perspective to do what I'm trying to do, but I haven't been able to figure out how to do it using the API. Am I missing something?
Edit: I didn't realize this was important, but I'm using the flutterfire firestore library.
A few points. It is ALWAYS a good practice to use random, well distributed documentId's in firestore for scale and efficiency. Related to that, there is effectively NO WAY to query by documentId - and in the few circumstances you can use it (especially for a range, which is possible but VERY tricky, as it requires inequalities, and you can only do inequalities on one field). IF there's a reason to search on an ID, yes it is PERFECTLY appropriate to store in the document as well - in fact, my wrapper library always does this.
the correct notation, btw, would be FieldPath.documentId() (method, not constant) - alternatively, __name__ - but I believe this only works in Queries. The reason it requested a new index is without the () it assumed you had a field named FieldPath with a subfield named documentid.
Further: FieldPath.documentId() does NOT generate the documentId at the server - it generates the FULL PATH to the document - see Firestore collection group query on documentId for a more complete explanation.
So net:
=> documentId's should be as random as possible within a collection; it's generally best to let Firestore generate them for you.
=> a valid exception is when you have ONE AND ONLY ONE sub-document under another - for example, every "user" document might have one and only one "forms of Id" document as a subcollection. It is valid to use the SAME ID as the parent document in this exceptional case.
=> anything you want to query should be a FIELD in a document,and generally simple fields.
=> WORD TO THE WISE: Firestore "arrays" are ABSOLUTELY NOT ARRAYS. They are ORDERED LISTS, generally in the order they were added to the array. The SDK presents them to the CLIENT as arrays, but Firestore it self does not STORE them as ACTUAL ARRAYS - THE NUMBER YOU SEE IN THE CONSOLE is the order, not an index. matching elements in an array (arrayContains, e.g.) requires matching the WHOLE element - if you store an ordered list of objects, you CANNOT query the "array" on sub-elements.
From what I've found:
FieldPath.documentId does not match on the documentId, but on the refPath (which it gets automatically if passed a document reference).
As such, since the documents are to be sorted by timestamp, it would be more ideal to create a timestamp fieldvalue for createdAt rather than a human-readable string which is prone to string length sorting over the value of the string.
From there, you can simply sort by date and limit to last. You can keep the document ID's as you intend.
I don't want to have automatic indexes created by firestore because I need to remove and add every five minutes 50-100 documents (each doc has +/-60 fields) to my subcollection. This causes of big volume for "Cloud Firestore Index Write Ops" (300k / day for only one user) and Cloud Storage. I don't need to sort, filtering that documents so I suppose I can turn off automatic indexes, right?
I know that I can add exemptions for fields, but I don't know how can I use it for documents in subcollections. What should I pass in Collection ID and Field path if the path for documents is like:
mainCollectionName/{id}/subcollectionName/{document=**}
and when should I select a collection checkbox and when collection group checkbox?
Unfortunately, it's not possible to disable indexes or create exemptions for documents to be indexed. As clarified in this similar post here, this cannot be achieved and there is even a limit of 200 exemptions of fields that can be done - you can check the limits here.
For your case, indeed, you would have to exempt the fields individually and besides that, to create the exemption, to set the collection you use its id and not the path. So, you would only need to set in the Collection ID field the subcollectionName and then the field to be exempted.
In addition to this, feel free to raise a Feature Request in Google's Issue Tracker, so they can check about implementing an exemption of documents in the future.
I have an Orders collection. It contains a field called venueId. And I'm querying against this field using isEqualTo. The venueId is the firebase user uid. I also have a venues collection. It contains this venueId and also has a list of VenueAdmins ids(These ids are also firebase user uids )The app is a point of sales app(pos). I need to query the orders collections so that valueAdmins and venueId see the correct stream. Is quite easy to query with venueId.. venueId,isEqualto, uid. I'm wondering what's the best approach to allow the venueAdmins see the stream as well.
|-Orders // collection
order. //doc
venueId:'2344567788999999'
|-Venues // collection
venue. //doc
venueAdmin: ['3333333333333','55555555555555555']
venueId:'2344567788999999'
My query builder so far: queryBuilder: (query) => query.where('venue.id', isEqualTo: uid)
Firestore does not have the capability to "join" documents from different collections in a single query. A single query can only consider documents in single collection at a time. The way you have your data structured now, it will require at least two queries. First, to find a venue, then second, to find the orders for an admin in a venue.
The only way to make this easier from the perspective of queries is to denormalize your data by duplicating venue data into the order documents. If each order also had a list of admins, then you could reduce this down to a single query.
Let's say I've got 100 documents in my collection. My goal is to get the documents between 5 and 33.
I tried startAt and endAt but it didn't work:
const db = firebase.firestore();
await db
.collection("pictures")
.startAt(start)
.endAt(end)
.get()
What you're trying to do isn't really possible. Documents don't have a natural index or position within a collection. To get ordering within a collection, you need to use at least one field on which you want to sort the documents. Only then do they have an order, and only then can you page through them.
The startAt and endAt methods on the query require that you define some order. You can see that in the example code provided in the API docs I linked to. Note the following statement in the docs for startAt:
Creates and returns a new Query that starts at the provided set of
field values relative to the order of the query. The order of the
provided values must match the order of the order by clauses of the
query.