Uber-style Company ChallengeEasyVerified answerSQLite live

Trips Without Ratings

Which completed trips have no rating record at all?

  • Joins
  • NULL handling
  • Filtering
  • Sorting

Challenge brief

Understand the request

Trust & Safety is investigating why some completed trips have no rating submitted — these need follow-up to maintain platform quality.

Find trips with no entry in the ratings table using an anti-join.

Return

  • trip_id
  • rider_name (full name)
  • driver_name (full name)
  • fare_amount

Constraints

  • Return only trips whose status is 'completed'
  • Exclude any trip that has a rating record
  • Order by trip ID

Data you will use

Review the relevant tables before deciding how to join, filter, or aggregate them.

trips

  • trip_idINTEGER
  • rider_idINTEGER
  • driver_idINTEGER
  • fare_amountREAL

riders

  • rider_idINTEGER
  • first_nameVARCHAR(50)
  • last_nameVARCHAR(50)

drivers

  • driver_idINTEGER
  • first_nameVARCHAR(50)
  • last_nameVARCHAR(50)

ratings

  • rating_idINTEGER
  • trip_idINTEGER

Hints, when you need them

Open one clue at a time so you still do the reasoning.

Hint 1

You need trips with NO matching row in ratings. This is an anti-join: LEFT JOIN trips to ratings, then keep only rows where the ratings side is NULL — meaning no rating was submitted.

Hint 2

INNER JOIN trips to riders and drivers for names. LEFT JOIN ratings on trip_id. WHERE rt.rating_id IS NULL keeps only unrated trips.

Hint 3

Preserve completed trips while optionally matching ratings, then keep only rows where the rating-side key is absent.

Verified SQL answer

Attempt the problem first, then compare structure and reasoning—not just syntax.

Reveal solution and explanation
SELECT t.trip_id, r.first_name || ' ' || r.last_name AS rider_name, d.first_name || ' ' || d.last_name AS driver_name, t.fare_amount FROM trips t INNER JOIN riders r ON t.rider_id = r.rider_id INNER JOIN drivers d ON t.driver_id = d.driver_id LEFT JOIN ratings rt ON t.trip_id = rt.trip_id WHERE t.trip_status = 'completed' AND rt.rating_id IS NULL ORDER BY t.trip_id

Why this works

Trips 1001-1004 all have rating rows; 1005 and 1006 do not. LEFT JOIN keeps all trips rows — unmatched ones have NULL in all ratings columns. WHERE rating_id IS NULL isolates those two unrated trips.

Success check

2 trips — trip 1005 (Emma Evans / David Brown, $35) and trip 1006 (Alice Anderson / John Doe, $22)

Expected result

Use this output to verify values, aliases, ordering, and row count.

trip_idrider_namedriver_namefare_amount
1005Emma EvansDavid Brown35
1006Alice AndersonJohn Doe22

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