I have the following documents:
{'variations': ['BlueViolet', 'CadetBlue', 'Cyan']}
{'variations': ['LightPink', 'VioletRed']}
And I want to write a query that selects all documents where size of intersection between variations field and {'Cyan', 'CadetBlue', 'SmoothsRed'} is greater than 2.
Can this be performed with mongodb operators?
This link explains how to achieve what you want - https://docs.mongodb.com/manual/reference/operator/aggregation/setIntersection/
In order to write the comparison, I'll have to assume each document can be uniquely identified using an _id value, and then you can write your query using the solution given in this answer - How to find set intersection of sets between the documents in a single collection in MongoDB?
Good luck
Related
I am trying to sort the result of my mongoDB query and add a ranking based on that sorting. Currently I only call .find().sort({total: 1}) and this gives me the correct ordering of the documents. But is it possible to "add a field" based on that sorting (basically a ranking field, starting from 1 and counting up)? I tried googling but didnt found anything that suits for this purpose.
Thanks in advance.
I have a mongodb collection called employeeInformation, in which I have two documents:
{"name1":"tutorial1"}, {"name2":"tutorial2"}
When I do db.employeeInformation.find(), I get both these documents displayed. My question is - is there a query that I can run to confirm that the collection contains only those two specified documents? I tried db.employeeInformation.find({"name1":"tutorial1"}, {"name2":"tutorial2"}) but I only got the id corresponding to the first object with key "name1". I know it's easy to do here with 2 documents just by seeing the results of .find(), but I want to ensure that in a situation where I insert multiple (100's) of documents into the collection, I have a way of verifying that the collection contains all and only those 100 documents (note I will always have the objects themselves as text). Ideally this query should work in mongoatlas console/interface as well.
db.collection.count()
will give you number of inserts once you have inserted the document.
Thanks,
Neha
QUERYING MONGODB: RETREIVE SHOPS BY NAME AND BY LOCATION WITH ONE SINGLE QUERY
Hi folks!
I'm building a "search shops" application using MEAN Stack.
I store shops documents in MongoDB "location" collection like this:
{
_id: .....
name: ...//shop name
location : //...GEOJson
}
UI provides to the users one single input for shops searching. Basically, I would perform one single query to retrieve in the same results array:
All shops near the user (eventually limit to x)
All shops named "like" the input value
On logical side, I think this is a "$or like" query
Based on this answer
Using full text search with geospatial index on Mongodb
probably assign two special indexes (2dsphere and full text) to the collection is not the right manner to achieve this, anyway I think this is a different case just because I really don't want to apply sequential filter to results, "simply" want to retreive data with 2 distinct criteria.
If I should set indexes on my collection, of course the approach is to perform two distinct queries with two distinct mehtods ($near for locations and $text for name), and then merge the results with some server side logic to remove duplicate documents and sort them in some useful way for user experience, but I'm still wondering if exists a method to achieve this result with one single query.
So, the question is: is it possible or this kind of approach is out of MongoDB purpose?
Hope this is clear and hope that someone can teach something today!
Thanks
I need to fetch the document in a mongodb collection using its position. I know the position of the document inside the collection exactly but could not figure out a way to pull those documents from collection. Is there any way to achieve this?
db.daily.find({'_id': {'$in': 0,5,8}})
This is what i tried but _id is not inserted as 1,2,3... but it has some random num Eg:57d8fd62f2a9d913ba0d006d. Thanks in advance.
You can use skip and limit to query based on the position in the natural order
db.collection.find().skip(10).limit(1) // get 10th document in natural order
As the natural order link points out, the document order need not match the order that documents are inserted (with an exception for capped collections). If you use the default ObjectId as the _id field for your documents you can sort by _id to order based on insertion in the collection (up to the resolution of the timestamp in the ObjectId)
db.collection.find().sort([("_id",1)]).skip(10).limit(1) // get 10th document in inserted order
You may also consider using your own _id or adding a field to be able to sort on in order to query based on the position you define.
I have a mongodb query, and I want to add a computed field. The computed field is based on where or not the item is in the results of another query. So my query returns the columns a,b,c,d, and then column e should be based on whether or not the current row would be matched by another query.
Is there an efficient way to do this in mongo? I'm not really sure how to do this one...
There is no way currently to execute a function as you describe within the database when returning a document via standard functions such as find. It's been requested by the community, but the general request is to operate only on a single document.
There are calculated fields using $project in the aggregation framework. But, they only operate on the current document in the pipeline. So, they can't summarize other queries.
You'll need to likely build your e value as part of your data access layer.