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Table: ParkingTransactions
+--------------+-----------+ | Column Name | Type | +--------------+-----------+ | lot_id | int | | car_id | int | | entry_time | datetime | | exit_time | datetime | | fee_paid | decimal | +--------------+-----------+ (lot_id, car_id, entry_time) is the primary key (combination of columns with unique values) for this table. Each row of this table contains the ID of the parking lot, the ID of the car, the entry and exit times, and the fee paid for the parking duration.
Write a solution to find the total parking fee paid by each car across all parking lots, and the average hourly fee (rounded to 2 decimal places) paid by each car. Also, find the parking lot where each car spent the most total time.
Return the result table ordered by car_id in ascending order.
Note: Test cases are generated in such a way that an individual car cannot be in multiple parking lots at the same time.
The result format is in the following example.
Example:
Input:
ParkingTransactions table:
+--------+--------+---------------------+---------------------+----------+ | lot_id | car_id | entry_time | exit_time | fee_paid | +--------+--------+---------------------+---------------------+----------+ | 1 | 1001 | 2023-06-01 08:00:00 | 2023-06-01 10:30:00 | 5.00 | | 1 | 1001 | 2023-06-02 11:00:00 | 2023-06-02 12:45:00 | 3.00 | | 2 | 1001 | 2023-06-01 10:45:00 | 2023-06-01 12:00:00 | 6.00 | | 2 | 1002 | 2023-06-01 09:00:00 | 2023-06-01 11:30:00 | 4.00 | | 3 | 1001 | 2023-06-03 07:00:00 | 2023-06-03 09:00:00 | 4.00 | | 3 | 1002 | 2023-06-02 12:00:00 | 2023-06-02 14:00:00 | 2.00 | +--------+--------+---------------------+---------------------+----------+
Output:
+--------+----------------+----------------+---------------+ | car_id | total_fee_paid | avg_hourly_fee | most_time_lot | +--------+----------------+----------------+---------------+ | 1001 | 18.00 | 2.40 | 1 | | 1002 | 6.00 | 1.33 | 2 | +--------+----------------+----------------+---------------+
Explanation:
- For car ID 1001:
- From 2023-06-01 08:00:00 to 2023-06-01 10:30:00 in lot 1: 2.5 hours, fee 5.00
- From 2023-06-02 11:00:00 to 2023-06-02 12:45:00 in lot 1: 1.75 hours, fee 3.00
- From 2023-06-01 10:45:00 to 2023-06-01 12:00:00 in lot 2: 1.25 hours, fee 6.00
- From 2023-06-03 07:00:00 to 2023-06-03 09:00:00 in lot 3: 2 hours, fee 4.00
- For car ID 1002:
- From 2023-06-01 09:00:00 to 2023-06-01 11:30:00 in lot 2: 2.5 hours, fee 4.00
- From 2023-06-02 12:00:00 to 2023-06-02 14:00:00 in lot 3: 2 hours, fee 2.00
Note: Output table is ordered by car_id in ascending order.
We can first group by car_id and lot_id to calculate the parking duration for each car in each parking lot. Then, we use the RANK() function to rank the parking duration of each car in each parking lot to find the parking lot where each car has the longest parking duration.
Finally, we can group by car_id to calculate the total parking fee, average hourly fee, and the parking lot with the longest parking duration for each car.
# Write your MySQL query statement below
WITH
T AS (
SELECT
car_id,
lot_id,
SUM(TIMESTAMPDIFF(SECOND, entry_time, exit_time)) AS duration
FROM ParkingTransactions
GROUP BY 1, 2
),
P AS (
SELECT
*,
RANK() OVER (
PARTITION BY car_id
ORDER BY duration DESC
) AS rk
FROM T
)
SELECT
t1.car_id,
SUM(fee_paid) AS total_fee_paid,
ROUND(
SUM(fee_paid) / (SUM(TIMESTAMPDIFF(SECOND, entry_time, exit_time)) / 3600),
2
) AS avg_hourly_fee,
t2.lot_id AS most_time_lot
FROM
ParkingTransactions AS t1
LEFT JOIN P AS t2 ON t1.car_id = t2.car_id AND t2.rk = 1
GROUP BY 1
ORDER BY 1;