Reputation: 10629
I am trying to create a summary with the best score of each student per day.
Example data:
╔════╦════════════╦═══════╦═══════════════════╗
║ id ║ student_id ║ score ║ date_time ║
╠════╬════════════╬═══════╬═══════════════════╣
║ 1 ║ 1 ║ 5 ║ 2018-07-01 9:30 ║
║ 2 ║ 1 ║ 3 ║ 2018-07-01 15:30 ║
║ 3 ║ 1 ║ 7 ║ 2018-07-02 8:30 ║
║ 4 ║ 2 ║ 7 ║ 2018-07-01 9:30 ║
║ 5 ║ 2 ║ 8 ║ 2018-07-01 15:30 ║
║ 6 ║ 2 ║ 8 ║ 2018-07-02 8:30 ║
║ 7 ║ 3 ║ 4 ║ 2018-07-02 10:30 ║
║ 8 ║ 3 ║ 10 ║ 2018-07-02 13:45 ║
╚════╩════════════╩═══════╩═══════════════════╝
Desired result:
╔════════════╦════════════╦═════════════╦═════════════╦══════════════════════╗
║ student_id ║ date ║ score_total ║ best_score ║ best_score_date_time ║
╠════════════╬════════════╬═════════════╬═════════════╬══════════════════════╣
║ 1 ║ 2018-07-01 ║ 8 ║ 5 ║ 2018-07-01 9:30 ║
║ 1 ║ 2018-07-02 ║ 7 ║ 7 ║ 2018-07-02 8:30 ║
║ 2 ║ 2018-07-01 ║ 15 ║ 8 ║ 2018-07-01 15:30 ║
║ 2 ║ 2018-07-02 ║ 8 ║ 8 ║ 2018-07-02 8:30 ║
║ 3 ║ 2018-07-02 ║ 14 ║ 10 ║ 2018-07-02 13:45 ║
╚════════════╩════════════╩═════════════╩═════════════╩══════════════════════╝
This is what I got so far (and my question):
SELECT
student_id,
date_time::date as date,
SUM(score) as score_total,
MAX(score) as best_score,
date_time as best_score_date_time -- << HOW TO GET THIS ONE?
FROM
score
GROUP BY
student_id,
date_time::date
Upvotes: 0
Views: 65
Reputation: 244
Below is the simple solution.
Replace
date_time as best_score_date_time
with
( array_agg(date_time order by score desc) )[1]
Upvotes: 5
Reputation: 13524
SELECT A.student_id,
A.date,
A.score_total,
A.score AS best_score,
A.best_score_date_time
FROM
(
SELECT student_id,
date_time::date AS date,
SUM( score ) OVER ( PARTITION BY date_time::date ) AS score_total,
score,
RANK( ) OVER ( PARTITION BY date_time::date ORDER BY score DESC ) AS rnk,
date_time
FROM score
) A
WHERE A.rnk = 1;
Upvotes: 2