C100DEV CRUD Operations Practice Question

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

• ✗

  db.shipments.deleteOne({ status: "cancelled" })

  Why it's wrong here

  deleteOne() removes only the first matching document, leaving all remaining cancelled shipments in place. Since the requirement is to clear every cancelled record, a single deletion would silently leave data behind. It is appropriate only when exactly one document should be removed, which is not the case here.

• ✓

  db.shipments.deleteMany({ status: "cancelled" })

  Why this is correct

  deleteMany() removes every document matching the filter, so passing the equality filter { status: "cancelled" } deletes all cancelled shipments in one operation. It returns a result document with the deletedCount, confirming how many records were removed. This is the direct, idiomatic way to purge a subset of documents based on a field value.

• ✗

  db.shipments.remove({ status: "cancelled" })

  Why it's wrong here

  The remove() method was deprecated in MongoDB 3.x and removed in the modern 5.0+ shell, so it would fail against current servers. Even where it existed, the supported replacement is deleteMany() with the same filter. In this scenario the command would not reliably execute, making it the wrong choice for a current deployment.

• ✗

  db.shipments.drop({ status: "cancelled" })

  Why it's wrong here

  drop() removes an entire collection and its indexes; it does not accept a filter argument in the way implied. Passing a query document would not selectively delete cancelled shipments and would instead affect the whole collection if the call were valid. It is far too destructive for a targeted status-based purge.