CODEWITHSUNDEEP
mongodb
user1409708
user1409708

Reputation: 1053

return only a single property "_id"

I would like to know if there is a way to return only the _id, user_id and total without the items subdocument.

{
    "_id" : 122,
    "user_id" : 123456,
    "total" : 100,
    "items" : [
            {
                    "item_name" : "my_item_one",
                    "price" : 20
            },
            {
                    "item_name" : "my_item_two",
                    "price" : 50
            },
            {
                    "item_name" : "my_item_three",
                    "price" : 30
            }
    ]
}

Upvotes: 35

Views: 38144

Answers (2)

radtek
radtek

Reputation: 36300

What I was looking for is returning a list of ids, so return only _id but as an array for all documents. I did it as such:

db. mycollection.distinct("_id", {})

The first parameter is your distict fields, and second is your query. More on distinct here. Hopefully this helps someone with a similar requirement. Cheers!

Upvotes: 31

McGarnagle
McGarnagle

Reputation: 102763

The second parameter of find lets you select fields. So you can use this (note that the _id field is always selected anyway):

db.mycollection.find({}, {"user_id": 1, "total": 1});

You can also exclude certain fields, so this would be equivalent:

db.mycollection.find({}, {"items": 0});

You can exclude _id field by doing:

db.mycollection.find({}, {"user_id": 1, "_id": 0});

Upvotes: 43

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