Table Schema
Inspect
Table
| Column | Type |
|---|---|
| trip_id | integer |
| driver_id | integer |
| city | varchar |
| fare | decimal |
Driver
Earnings
Ranking
Rank drivers by total earnings within each city.
Inspect
Sample Data
Input
Output
Sample Input: trips
| trip_id | driver_id | city | fare |
|---|---|---|---|
| 1 | 201 | NYC | 20 |
| 2 | 202 | NYC | 30 |
| 3 | 201 | NYC | 25 |
| 4 | 203 | SF | 40 |
| 5 | 204 | SF | 35 |
| 6 | 203 | SF | 20 |
Expected Output
| driver_id | city | rank |
|---|---|---|
| 201 | NYC | 1 |
| 202 | NYC | 2 |
| 203 | SF | 1 |
| 204 | SF | 2 |
SQL Editor
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Query
Waiting for query
| driver_id | city | rank |
|---|---|---|
| 201 | NYC | 1 |
| 202 | NYC | 2 |
| 203 | SF | 1 |
| 204 | SF | 2 |
Hints
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Clues
Hint 01: Identify the grouping level required by the output.
Hint 02: Aggregate with COUNT, SUM, AVG, or a window function as needed.
Hint 03: Filter after aggregation with HAVING or after ranking with an outer query.
Solution
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Answer
Solution is locked until you decide to reveal it. Try the editor first, then open this when you want the reference answer.
WITH earnings AS ( SELECT driver_id, city, SUM(fare) AS total_earnings FROM trips GROUP BY driver_id, city ) SELECT driver_id, city, RANK() OVER (PARTITION BY city ORDER BY total_earnings DESC) AS rank FROM earnings;
Explanation
Step By
Step
01
Read the expected output columns to determine the final grain.
02
Aggregate or rank the input rows to calculate the requested metric.
03
Filter, sort, and alias the final columns to match the output.
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