I want to query by date and status. I created an index for both the fields, but when querying it throws an error:
Cannot have inequality filters on multiple properties: [created_at, status]
const docQuery = db.collection('documents')
.where('created_at', '<=', new Date()) // ex. created_at later than 2 weeks ago
.where('status', '!=', 'processed') // works when == is used
.limit(10);
'status' is a string and can be multiple things.
I read about query limitations, https://firebase.google.com/docs/firestore/query-data/queries but that is such a simple thing for any database to do, what is a good solution for such a problem? Loading entire records is not an option.
The issue isn't that it's a "simple" thing to do. The issue is that it's an unscalable thing to do. Firestore is fast and cheap because of the limitations it places on queries. Limiting queries to a single inequality/range filter allows it to scale massively without requiring arbitrarily large amounts of memory to perform lots of data transformations. It can simply stream results directly to the client without loading them all into enough memory to hold all of the results. While it's not necessary to understand how this works, it's necessary to accept the limitations if you want to use Firestore effectively.
For your particular case, you could make your data more scalable to query by changing to your "status" field from a single string field to a set of boolean fields. For each one of the possible status values, you could instead have a boolean field that indicates the current status. So, for the query you show here, if you had a field called "isProcessed" with a boolean value true/false, you could find the first 10 unprocessed documents created before the current date like this:
const docQuery = db.collection('documents')
.where('created_at', '<=', new Date())
.where('processed', '==', false)
.limit(10);
Yes, this means you have to keep each field up to date when there is a change in status, but you now have a schema that's possible to query at massive scale.
Related
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.
This is a follow-up/elaboration to a previous question of mine.
In the case of a collection of documents containing a time range represented by two timestamp fields (start and end), how does one go about guaranteeing that two documents don't get added with overlapping time ranges?
Say I had the following JavaScript on form submit:
var bookingsRef = db.collection('bookings')
.where('start', '<', booking.end)
.where('end', '>', booking.start);
bookingsRef.get().then(snapshot => {
// if a booking is found (hence there is an overlap), display error
// if booking is not found (hence there is no overlap), create booking
});
Now if two people were to submit overlapping bookings at the same time, could transactions be used (either on the client or the server) to guarantee that in between the get and add calls no other documents were created that would invalidate the original collection get query where clauses.
Or would my option be using some sort of security create rule that checks for other document time overlaps prior to allowing a new write (if this is at all possible)? One approach to guarantee document uniqueness via security rules seems to be exposing field values in the document ID, but I'm not entirely sure how exposing the start and end timestamp values in the ID would allow a rule to check for overlapping time ranges.
I think transaction is proper approach. According to the documentation:
..., if a transaction reads documents and another client
modifies any of those documents, Cloud Firestore retries the
transaction. This feature ensures that the transaction runs on
up-to-date and consistent data.
This seems to be an answer to your problem. All reads will be retried, if anything will change in the meantime. I think transaction mechanism is exactly for that reason.
I have the following structure in a Firestore collection. The "ranks" collection is updated with documents named after the timestamps. In each document, I have the same fields and values. How can I query all documents for a specific field without parsing the entire document? I.e. I want all values in all documents where field is "aave"?
I am new to Firestore and I've been trying this for several weeks now. I tried limiting with where and considered using sub collection group queries but in my case data is not stored in sub collections. Sorry, for not being able to provide more context, since I couldn't get much closer.
Queries select specific values, or ranges of values, of a known field. There is no support for dynamic field names in a query in Firestore.
But if you want to get all documents where the field aave exists/has any value, you can make use of the fact that in the sort order of values they always start with null. So to get all documents where the field aave exists/has any value, you could do:
firebase.firebase().collection("ranks").where("aave", ">=", null)
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'm struggling to make a (not so) complex query in firebase cloud firestore.
I simply need to get all docs where the id field == a given id.
Then order the results by their date and limit to 10 results.
So this is my current situation
db.firestore().collection('comments')
.where("postId",'==',idOfThePost)
.orderBy('date','asc').limit(10).get().then( snapshot => {
//Nothing happens and the request wasn't even been executed
})
I can get the result only if i don't use the orderBy query but i have to process this sorting for the needs of my application.
Someone has an idea to help me to fix this ?
thanks
You can do this by creating an Index in the firestore.
The first field of the index should be the equality field and the second field of the index should be the order by field.
Given your problem, you would need the following index:
first field: postId, 'asc'
second field: date, 'asc'
Please check the doc. It says
However, if you have a filter with a range comparison (<, <=, >, >=), your first ordering must be on the same field
you can try this code
db.firestore().collection('comments')
.where("postId",'==',idOfThePost)
.orderBy('postId')
.orderBy('date','asc').limit(10).get().then( snapshot => {
.....
})
My Workaround
If you're googling this you've probably realized it can't be done traditionally. Depending on your problem though there may be some viable workarounds, I just finished creating this one.
Scenario
We have an app that has posts that appear in a feed (kind of like Reddit), each post has an algorithmic score 'score' and we needed a way to get the 'top posts' from 12-24 hours ago. Trying to query sorted by 'score' where timestamp uses > and < to build the 12-24 hour ago range fails since Firebase doesn't allow multiple conditional querying or single conditional querying with an descending sort on another field.
Solution
What we ended up doing is using a second field that was an array since you can compound queries for array-contains and descending. At the time a post was made we knew the current hour, suppose it was hour 10000 since the server epoch (i.e. floor(serverTime/60.0/60.0)). We would create an array called arrayOfHoursWhenPostIsTwelveToTwentyFourHoursOld and in that array we would populate the following values:
int hourOffset = 12;
while (hourOffset <= 24) {
[arrayOfHoursWhenPostIsTwelveToTwentyFourHoursOld addObject:#(currentHour+hourOffset)];
hourOffset++;
}
Then, when making the post we would store that array under the field hoursWhenPostIsTwelveToTwentyFourHoursOld
THEN, if it had been, say, 13 hours since the post was made (the post was made at hour 10000) then the current hour would be 10013, so we could use the array-contains query to see if our array contained the value 10013 while also sorting by algorithm score at the same time
Like so:
FIRFirestore *firestore = [Server sharedFirestore];
FIRCollectionReference *collection = [firestore collectionWithPath:#"Posts"];
FIRQuery *query = [collection queryOrderedByField:#"postsAlgorithmScore" descending:YES];
query = [query queryWhereField:#"hoursWhenPostIsTwelveToTwentyFourHoursOld" arrayContains:#(currentHour)];
query = [query queryLimitedTo:numberToLoad];
Almost Done
The above code will not run properly at first since it is using a compound index query, so we had to create a compound index query in firebase, the easiest way to do this is just run the query then look at the error in the logs and firebase SDK will generate a link for you that you can navigate to and it will auto-generate the compound index for your database for you, otherwise you can navigate to firebase>database>index>compound>new and build it yourself using hoursWhenTwelveToTwentyFourHoursOld: Arrays, score: Descending
Enjoy!
same here, it is weird why can't. below is another sample. can't get the result. Hoping firebase can reply about this and update the document.
dbFireStore.collection('room').where('user_id.'+global.obj_user.user_id,'==',true).orderBy('last_update').get().then((qs)=>{
console.log(qs);
});
using other work-around solution is javascript array and array.sort()
I ran into the same issue yesterday on Android. The Callback was just not called. Today I suddenly got an error message. FAILED_PRECONDITION: The query requires an index. It even contains a URL in the error message to generate that index with one click.
It seems that if you want to use orderBy on your data, you need to create an index for that field. The index also needs to be in the correct order (DESC, ASC).
As per firestore document,
If you attempt a compound query with a range clause that doesn't map to an existing index, you receive an error. The error message includes a direct link to create the missing index in the Firebase console.
So just click that link you get in Logcat, it will be redirected to create index page, just create index. It will take some time. after enabling composite index, you will get the result as your requested query.
Stumbled across this looking for help when i found that using the orderBy function didnt work and the documentation still says it does not support it. A bit weird and unclear to be honest, because it does support it so long as you index your Firestore database. For example, this query now works fine for me having set up indexing:
const q = query(docRef, where("category", "==", 'Main'), orderBy('title', 'asc')
Indexing in Firestore
Console Log that even gives you the url to automatically create the index if you try and run with the above command.
Maybe I am missing something, or a later version of Firebase (I am using v9) simply does support it.