[英]how do I group by day and 2 hrs in postgresql with a “time” field?
我需要按照一天的時間進行分組,然后將其分為2個小時:
我已經使用了EXTRACT PostgreSQL函數。 但是無法找到一種方法來按2小時的時間分組
SELECT EXTRACT(dow from completed_at) AS "day",
EXTRACT(hour from completed_at) AS "hour", count(*)
FROM orders
WHERE completed_at is not null
GROUP BY 1, 2
ORDER BY 1;
預期產量:
day hour count
-------- ------ ------
Sun 12am 10
Sun 2am 8
Sun 4am 0
Sun 6am 24
Sun 8am 25
Sun 10am 100
Sun 12pm 67
Sun 2pm 10
Sun 4pm 10
Sun 6pm 10
Sun 8pm 10
Sun 10pm 10
這樣的話,我平日都需要
嘗試:
SELECT EXTRACT(dow from completed_at) AS "day",
EXTRACT(hour from completed_at) AS "hour", count(*)
FROM orders
join generate_series(0,22,2) g on g >= extract(hour from completed_at) and g< extract(hour from completed_at) + 2
WHERE completed_at is not null
GROUP BY "day","hour"
ORDER BY 1;
就像在我的示例架構中一樣:
db=# create table so (t timestamptz);
CREATE TABLE
Time: 171.144 ms
db=# insert into so select generate_series(now(),current_date + 2,' 1hour'::interval);
INSERT 0 40
Time: 71.150 ms
db=# select count(*), g
from so
join generate_series(0,22,2) g on g >= extract(hour from t) and g< extract(hour from t) + 2
group by g
order by 2,1
;
count | g
-------+----
1 | 0
2 | 2
2 | 4
2 | 6
3 | 8
4 | 10
4 | 12
4 | 14
4 | 16
4 | 18
4 | 20
4 | 22
2 | 24
(13 rows)
Time: 11.958 ms
我使用紀元(秒),然后轉換為2小時
SELECT EXTRACT(dow from completed_at) AS "Day",
EXTRACT(hour from (to_timestamp(floor((extract('epoch' from completed_at) / (60 * 60 * 2) )) * (60 * 60 * 2))
AT TIME ZONE 'UTC')),
COUNT(*)
FROM orders
WHERE completed_at is not null
GROUP BY 1, 2
ORDER BY 1;
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