1

I have two collections: collection A and collection B. Collection B "A_id" will have the _id from Collection A (kinda like a relationship).

Collection A

{
  _id,
  name
}

Collection B

{
 _id,
 A_id,
 other_id
}

I need to sort the documents from Collection A by the amount of repetitions of documents to an A_id in Collection B (sort by count if you may). The constraint is that we don't want to have an array in collection A because the documents in B will grow a lot and will be updated constantly.

Any idea how to perform this kind of sorting?

Thanks much

Luis Porras
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2 Answers2

2

There are many ways to do what you're asking. You have to know some things about the data first. How many documents will a B document have?

  • It looks like a case where you'd want Collection A to have an items field with documents that are B. There is a 16 MB limit for that, but for most cases works just fine.

  • Keep a count on A with numbers of B records inserted. (Ie every time you insert a B you'd $inc the field on A). You'd then do an index on the count field for A and do sort({count: -1})

Justin Thomas
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1

It is my understanding that this is not strictly possible. However, you could use the aggregation framework group function and $sum accumulator to group by A_id which will give you a mapping between A_id and the total number of documents in collection B with that A_id. Sort that mapping and then you can select things from collection A in the order you have described, although you'd have to select from collection A one at a time since you are enforcing an ordering at selection time.

The other option would be to have some count field in collection A on each document that you include logic to increment every time something is inserted into in collection B with a particular A_id.

chrisdrobison
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  • The first option just inspired me on a similar problem at hand since I can't change data schema. A little improvement: we don't have to select one at a time. We can sort the result selected from A by the order of `A_id`s we just calculated with B. – JJPandari Sep 06 '16 at 09:43