MongoDB $unwind

One document in, one document per array element out — which is the whole operator, and also the whole problem.

11 lessons5 exercises4 medium · 1 hard

$unwind flattens an array field. A document whose items array holds three entries becomes three documents, identical except that each one carries a single items object instead of the array.

Run this and read the row count rather than the rows:

db.orders.aggregate([
  { $match: { _id: 1 } },
  { $unwind: "$items" },
  { $project: { _id: 1, userId: 1, items: 1 } }
])

One order went in. More than one came out, and each result still claims _id: 1 — because $unwind does not create new orders, it creates new rows about one order.

That is the source of nearly every $unwind bug:

Counts after an $unwind are counts of line items, not of orders. { $sum: 1 } following an unwind answers “how many item entries”, which is a perfectly good question and almost never the one that was asked. If you want orders, group back by $_id first, or do not unwind at all.

Sums need the multiply. After unwinding, each row holds one item, so revenue is price × quantity per row and then summed. Summing price alone is the classic slip, and it silently returns a smaller, entirely plausible number.

An empty array makes the document vanish. No elements, no rows. preserveNullAndEmptyArrays: true keeps it, which matters after a $lookup that found nothing.

The pairing with $group is so common that it has its own module below. The exercises make you decide which side of the unwind the grouping belongs on.