MongoDB: Update field with size of embedded array

Question

I have a collection of documents with an array (set in this case): my_array

I'm adding things to this set periodically

collection.find({ "_id": target_id })
.upsert().update({ '$addToSet':{ 'my_array': new_entry }})

Many of the logical operations I perform on this DB are based on this sub-array's size. So I've created a field (indexed) called len_of_array. The index is quite critical to my use case.

In the case where this is a true array and not a set, the $incr would work beautifully in the same update

However, since the sub-collection is a set, the length of the collection, my_array, may or may not have changed.

My current solution:

Call this periodically for each target_id, but this requires performing a find in order to get the correct len_of_array

collection.find({ '_id': target_id})
.upsert().update({ '$set':{ 'len_of_array': new_length }})

My Question Is there a way to set a field of a document to the indexed size of a sub-array in the same document in a single update?

Markus W Mahlberg · Accepted Answer · 2016-01-08 10:44:39Z

2

You don't need the field length_of_array in order to query by its size. There is the $size operator. Which would save you the periodical update, too.

Let's say you want to find all documents for which the length of my_array is greater than 2:

db.coll.find({ "my_array":{ "$size" :{ "$gt": 2 }}})

answered Jan 8, 2016 at 10:44

Markus W Mahlberg

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4 Comments

puj Over a year ago

Thanks. Am I correct in assuming that there is no way to index $size I will update my question to explain that the field len_of_array is also indexed.

Markus W Mahlberg Over a year ago

@puj Simply index the array.

mils Over a year ago

Is this correct? My recent experience suggests that indexing an array (e.g. putting an index on a "doc.arr" field) does not allow for indexed count, but rather the count will scan the document. Whereas searching for arrays that contain the value "x" will use the index.

Markus W Mahlberg Over a year ago

@mils Curious. Will check that. I think I had checked it when I wrote the answer, but I am not too sure.

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