I have a scraped some sports data, which contains an id field that ideally I'd like to use as Mongo's "_id". It's event based data and the IDs are meant to be unique however there is a few cases where they are not. However each entry also has a match id value, which is unique. There is never duplicated event IDs for the same match id. I was looking for a way to concatenate the two fields to create a unique id value that I could then use as the "_id" but this doesn't seem possible. I tried using the $concat operation with $tostring however this results in values such as 54821.4e+09
For example, a document with match id 3181621 and event id 1339018347 using
{$project:{"new_id":{$concat:[{"$toString":"$matchId"},"",{"$toString":"$id"}]}}}
results in "new_id" : "3181621.33902e+09", when I would want it to be 31816211339018347
I would like to check if a certain value is present in my Cloud Firestore collection through all the present fields and have back the document ID that has at least one field whose value is the one searched.
In this example, the code should give back only 2 records when I look for "Peter": 8cyMJG7uNgVoenA63brG and fnk0kgW7gSBc3EdOYWxD.
I know how to do a search when the field name is known. But in this case, I cannot know the field name at prior.
If you don't know the name of a field, you can't perform any queries against its value. Firestore requires queries to use some index, and indexes always work with the names of fields in your documents.
I have been trying to figure out a way to query a list of documents where I have a range filter on one field and order by another field which of course isn't possible, see my other question: Order by timestamp with range filter on different field Swift Firestore
But is it possible to save documents with the timestamp as id and then it would sort by default? Or maybe hardcode an ID, then retrieve the last created document id and increase id by one for the next post to be uploaded?
This shows how the documents is ordered in the collection
Any ideas how to store documents so they are ordered by created at in the collection?
It will order by document ID (ascending) by default in Swift.
You can use .order(by: '__id__') but the better/documented way is with FieldPath documentID() I don't really know Swift but I assume that it's something like...
.order(by: FirebaseFirestore.FieldPath.documentID())
JavaScript too has an internal variable which simply returns __id__.
.orderBy(firebase.firestore.FieldPath.documentId())
Interestingly enough __name__ also works, but that sorts the whole path, including the collection name (and also the id of course).
If I correctly understood your need, by doing the following you should get the correct order:
For each document, add a specific field of type number, called for example sortNbr and assign as value a timestamp you calculate (e.g. the epoch time, see Get Unix Epoch Time in Swift)
Then build a query sorted on this field value, like:
let docRef = db.collection("xxxx")
docRef.order(by: "sortNbr")
See the doc here: https://firebase.google.com/docs/firestore/query-data/order-limit-data
Yes, you can do this.
By default, a query retrieves all documents that satisfy the query in
ascending order by document ID.
See the docs here: https://firebase.google.com/docs/firestore/query-data/order-limit-data
So if you find a way to use a timestamp or other primary key value where the ascending lexicographical ordering is what you want, you can filter by any fields and still have the results sorted by the primary key, ascending.
Be careful to zero-pad your numbers to the maximum precision if using a numeric key like seconds since epoch or an integer sequence. 10 is lexicographical less than 2, but 10 is greater than 02.
Using ISO formatted YYYY-mm-ddTHH:MM:SS date-time strings would work, because they sort naturally in ascending order.
The order of the documents shown in the Firebase console is mostly irrelevant to the functioning of your code that uses Firestore. The console is just for browsing data, and that sorting scheme makes it relatively intuitive to find a document you might be looking for, if you know its ID. You can't change this sort order in the console.
Your code is obviously going to have other requirements, and those requirements should be coded into your queries, without regarding any sort order you see in the dashboard. If you want time-based ordering of your documents, you'll have to store some sort of timestamp field in the document, and use that for ordering. I don't recommend using the timestamp as the ID of a document, as that could cause problems for you in the future.
I would like to encode some meaning behinds first N characters of every document ID i.e. make first three characters determine a document type sensible to the system being used in.
You can have a custom _id when you insert the document. If the document to be inserted doesn't contain _id, then MongoDB will insert a ObejctId for you.
The _id can be of any type but keeping it uniform for all the documents makes sense if you are accessing from application layer.
You can refer one of the old questions at SO - How to generate unique object id in mongodb
I have problem whit long Mongo find results. Example how can i start query starting from _id X
to forward Example I know I have document where is 1000 users details I know there is user called Peter in list I can make query Users.find({userName: "Peter"}) and get this on user _id but how I can get all users also after this with out I need return JSON from above "Peter"
With the little amount of information you have given, You need to do this in two steps:
Get the id of the first record that matches the name "peter".
db.test.findOne({"userName":"Peter"},{"_id":1});
Returns one document that satisfies the specified query criteria. If
multiple documents satisfy the query, this method returns the first
document according to the natural order which reflects the order of
documents on the disk. In capped collections, natural order is the
same as insertion order.
Once you have the id of the record with peter, you can retrieve the records with their id > the id of this record.
db.test.find({"_id":{$gte:x}});
Where, x is the id of the first record returned by the first query.