CODEWITHSUNDEEP
mongodbindexingmongodb-query
sme
sme

Reputation: 4163

Getting the position of indexed properties

Say I have a collection, with documents like:

{
    _id: 1,
    name: "Bob",
    skillA: 500,
    skillB: 1580,
    skillC: 750
}

and each the colleciton has the following indices:

db.mycollection.createIndex({ "skillA": -1 })
db.mycollection.createIndex({ "skillB": -1 })
db.mycollection.createIndex({ "skillC": -1 })

I want to be able to get the position of each index property in their respective index. For example, if skillA is at the first position in the index, then I would like to find a way to return 0 for that.

I understand if I wanted to get the position of the entire document in the collection, I'd just search for documents with an _id $lt whatever, (assuming _id is auto-incremented), and then just use .count(), but I don't know how to do something like this just for the index collection.

Upvotes: 0

Views: 34

Answers (1)

Joe
Joe

Reputation: 28356

You could leverage a covered query sorted by the field in question, which should use a COUNT_SCAN stage instead of fetching any documents:

db.mycollection.find({skillA: {$gt: 500}},{_id:0,skillA:1}).sort({skillA:-1}).count()

Upvotes: 1

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