Question

I have a document with only about 200 elements in it. But I am still getting a very slow response just by querying it like:

Product.order_by([:name, :asc]).limit(20)

I do have an index on name which I have verified exists in the mongo shell. Is the only way to return efficiently sorted results to change they _id value to something like

"#{name}_#{random_id}"

Answer 1

Short answer: absolutely not.
Long answer: let’s make sure the query is using your index (not just that the index exists).

Our query:

This does not use an index (in this case, because it does not exist)

db.products.find().sort( { name: -1}).explain()
{
    "cursor" : "BasicCursor",
    "nscanned" : 1042,
    "nscannedObjects" : 1042,
    "n" : 1042,
    "scanAndOrder" : true,
    "millis" : 5,
    "nYields" : 0,
    "nChunkSkips" : 0,
    "isMultiKey" : false,
    "indexOnly" : false,
    "indexBounds" : {
        }

Let’s prove it

db.products.getIndexes()
[
    {
        "name" : "_id_",
        "ns" : "project_development.products",
        "key" : {
            "_id" : 1
        }
    }
]

Create the index

db.products.ensureIndex({name:1});

Now when we explain the query:

db.products.find().sort( { name: -1}).explain()
{
    "cursor" : "BtreeCursor name reverse",
    "nscanned" : 1042,
    "nscannedObjects" : 1042,
    "n" : 1042,
    "millis" : 2,
    "nYields" : 0,
    "nChunkSkips" : 0,
    "isMultiKey" : false,
    "indexOnly" : false,
    "indexBounds" : {
        "name" : [
            [
                {
                    "$maxElement" : 1
                },
                {
                    "$minElement" : 1
                }
            ]
        ]
    }
}