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I would like to run following query:

Group.find({program: {$in: [...]}}).lean().select('_id')

And then NOT get following back:

[{_id: ...}, {_id: ...}, {_id: ...}, {_id: ...}]

BUT following:

[..., ..., ..., ...]

where ... represents an _id of a Group.

Of course I could just run the query and then loop through the Groups I get back, but I would like to do it in the query if possible, because that's probably going to be faster.

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2 Answers 2

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57 
Group.find({program: {$in: [...]}})
  .distinct('_id')

db.collection.distinct(field, query)

Finds the distinct values for a specified field across a single collection and returns the results in an array.

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3 Comments

Thank you, I will try that tomorrow!
Why do you have lean() in it the distinct will only return the ids?
@ElishaSterngold I agree, it's certainly redundant - thanks for noticing, I updated my answer to reflect that.
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You can also pass the query directly into a distinct() call:

Group.distinct('_id', {program: {$in: [...]}})

which would return an array of ObjectIds satisfying the query.

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