Reputation: 8318
If I have a mongodb collection users
like this:
{
"_id": 1,
"name": {
"first" : "John",
"last" :"Backus"
},
}
How do I retrieve name.first
from this without providing _id
or any other reference. Also, is it possible that pulling just the `name^ can give me the array of embedded keys (first and last in this case)? How can that be done?
db.users.find({"name.first"})
didn't work for me, I got a:
SyntaxError "missing: after property id (shell):1
Upvotes: 9
Views: 51589
Reputation: 119
In 2022
const cursor = db
.collection('inventory')
.find({
status: 'A'
})
.project({ item: 1, status: 1 });
Source: https://www.mongodb.com/docs/manual/tutorial/project-fields-from-query-results/
Upvotes: 1
Reputation: 11
To fetch all the record details: db.users.find({"name.first":""}) To fetch just the name or specific field: db.users.find({{},"name.X":""}); where X can be first, last .
dot(.) notation can be used if required to traverse inside the array for key value pair as db.users.find({"name.first._id":"xyz"});
Upvotes: 0
Reputation: 16335
The first argument to find()
is the query criteria whereas the second argument to the find()
method is a projection, and it takes the form of a document with a list of fields for inclusion or exclusion from the result set. You can either specify the fields to include (e.g. { field: 1 }
) or specify the fields to exclude (e.g. { field: 0 }
). The _id
field is implicitly included, unless explicitly excluded.
In your case, db.users.find({name.first}) will give an error as it is expected to be a search criteria.
To get the name json :
db.users.find({},{name:1
})
If you want to fetch only name.first
db.users.find({},{"name.first":1})
Mongodb Documentation link here
Upvotes: 31