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.updateOne({ _id: 42 }, { $pull: { packages: { trackingCode: "TRK-9981" } } })

  Why this is correct

  The $pull operator removes every array element that matches the supplied condition, so this deletes only the embedded package whose trackingCode equals TRK-9981 while preserving the rest of the array and the parent document. The filter on _id restricts the update to the intended shipment, making this the precise and minimal operation for the scenario.

• ✗

  db.shipments.updateOne({ _id: 42 }, { $pullAll: { packages: ["TRK-9981"] } })

  Why it's wrong here

  $pullAll removes all instances of the exact values listed from an array. Here the array holds embedded documents, not bare strings, so specifying the string "TRK-9981" matches nothing and the update changes no data. It cannot match on a field inside the subdocument, so it does not satisfy the scenario.

• ✗

  db.shipments.updateOne({ _id: 42, "packages.trackingCode": "TRK-9981" }, { $unset: { "packages.$": "" } })

  Why it's wrong here

  $unset with the positional operator sets the matched array element to null rather than deleting it, leaving a null placeholder inside the array. The document would still contain a packages entry, which violates the requirement to remove the subdocument entirely. It also depends on the positional operator matching the array element via the query predicate.

• ✗

  db.shipments.updateOne({ _id: 42 }, { $pop: { packages: 1 } })

  Why it's wrong here

  $pop removes either the first or last element of an array by position, not by matching a field value. With a value of 1 it deletes the final element of the packages array, which could be any package, not necessarily the one with trackingCode TRK-9981. It cannot express a value-based removal, so it fails the requirement.