I'm trying to find the id of a doc where a map value in an array of maps equals "x".
in the following case, I'm trying to find which rep owns the cause with code "hog"
I'll likely be going down the denormalizing route, but is this possible?
Firestore has an array-contains operator that you can use to query whether a certain item exists in an array field, but that operator only works if you specify the exact, complete value of the field. It can't test for a partial match.
The common approach to your use-case is to add an additional array field with just the values you want to query on, i.e.
cause-codes: ["hog"]
Once you modified your documents with this additional field, you can then use a query like:
repsRef.where('cause-codes:', 'array-contains', 'hof')
Related
im trying to make a sort of chat app with the following architecture:
Its currently working as follows: in a root doc, there is an array of chat objects. Each chat object spawns a message doc, which contains an array of message objects. Same logic for comments.
Im thinking about a function to update all posts relative to the associated user, IE if a user changes their name all associated comments will be updated. Is this possible with the WhereIn() function? Or should i edit the architecture to something more like each document being its own message/comment? Thanks!
An in clause checks if a specific field is equal to one of a set of values, which doesn't apply here.
You might be thinking of array-contains, but that wouldn't work in your current structure either. The array-contains only matches exact an item in the array if it completely/exactly matches the value in the query.
The common way to allow such a query is to add a participants array field to each document where you store the UIDs of all participants in that doc. Then you can do an array-contains query against that.
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 have question about query in firecloud.
For example, see my document:
i need filter all documents where the account_id is "value" and "approve" is true. How i do it?
I tried: where("members.account_id", "==, "qgZ564nfEaNP3Nt4cW1K3jCeVlY2") and not work:
See, this is a question that not for specific language, just about firestore query, after undestand, i go apply it in my code.
I'm programing in flutter.
If you query like this:
where("members.account_id", "==", "qgZ564nfEaNP3Nt4cW1K3jCeVlY2")
Firebase checks if there's a nested field /members/account_id with the given value. And such a field doesn't exist, because the value is actually at /members/0/account_id.
Since you don't know the position in the array the the ID exists, you instead want to use an array-contains condition. With array-contains Firestore checks if the array you contains the exact member that you specify.
With your current structure you can check if the member exists with:
.where("members", "array-contains", { accountId: "qgZ564nfEaNP3Nt4cW1K3jCeVlY2", approved: true })
So you must know the entire data of the array element in order to use array-contains.
If you only know a single member, while you're storing multiple members, you can't use array-contains. In that case, I recommend adding an additional member_accountIds field to each document, where you store just their account IDs:
member_accountIds: ["qgZ564nfEaNP3Nt4cW1K3jCeVlY2", ... ]
With that in place, you can query like this:
.where("members_accountIds", "array-contains", "qgZ564nfEaNP3Nt4cW1K3jCeVlY2")
I have the following document structure in firebase:
{
typeId: number,
tripId: number,
locationId: string,
expenseId: number,
createtAt: timestamp
}
I want to query this collection using different 'where' statement everytime. Sometimes user wants to filter by type id and sometimes by locationId or maybe include all of the filters.
But it seems like I would need to create a compound index of each possible permutation? For example: typeId + expenseId, typeId + locationId, location + expenseId, etc, otherwise it doesn't work.
What if I have 20 fields and I want to make it possible to search across all of these?
Could you please help me to construct a query and indexes for the following requirement: Possibility to query across all fields, query can contain one, two, three, all or no fields included in where clause and always has to be ordered descending order by createdAt.
Cloud Firestore automatically creates indexes for the individual fields of your documents. So it can already filter on each field without you have to manually add these indexes.
In many cases it is able to combine these indexes to allow queries on field combinations, by performing a so-called zig-zag-merge-join.
Custom additional indexes are typically only needed once you add an ordering-clause to your query, in addition to filter clauses. If you have such a case, the Firestore client will log an error telling you exactly what index to create (with a link to the Firestore console that is prepopulated to created the index for you).
Consider the following data in Firestore:
I also have a set of selected by the user he's interested in: ['python', 'sql', 'swift']
How to get all the documents from my collection where tags field contains at least one of tags selected by the user? I'm aware that Firestore doesn't have such operator out of the box. Probably I need to rethink the way I store data somehow? Please point me to the right direction.
With the way you have your data structured, you will need to make one array-contains type query for each of the tags the user is interested in, then merge the results of each of those queries in your app code.