[英]How to get generate_series() function in Redshift to result in just dates NOT date-time
[英]Handling of generate_series() in queries with date or timestamp with / without time zone
我有一個查詢來生成基於按date
和employee_id
分組的日期系列的報告。 日期應基於特定時區,在本例中為“Asia/Kuala_Lumpur”。 但這可能會根據用戶所在時區的位置而改變。
SELECT
d::date AT TIME ZONE 'Asia/Kuala_Lumpur' AS created_date,
e.id,
e.name,
e.division_id,
ARRAY_AGG(
a.id
) as rows,
MIN(a.created_at) FILTER (WHERE a.activity_type = 1) as min_time_in,
MAX(a.created_at) FILTER (WHERE a.activity_type = 2) as max_time_out,
ARRAY_AGG(
CASE
WHEN a.activity_type = 1
THEN a.created_at
ELSE NULL
END
) as check_ins,
ARRAY_AGG(
CASE
WHEN a.activity_type = 2
THEN a.created_at
ELSE NULL
END
) as check_outs
FROM (SELECT MIN(created_at), MAX(created_at) FROM attendance) AS r(startdate,enddate)
, generate_series(
startdate::timestamp,
enddate::timestamp,
interval '1 day') g(d)
CROSS JOIN employee e
LEFT JOIN attendance a ON a.created_at::date = d::date AND e.id = a.employee_id
where d::date = date '2020-11-20' and division_id = 1
GROUP BY
created_date
, e.id
, e.name
, e.division_id
ORDER BY
created_date
, e.id;
餐桌attendance
的定義和樣本數據:
CREATE TABLE attendance (
id int,
employee_id int,
activity_type int,
created_at timestamp with time zone NOT NULL
);
INSERT INTO attendance VALUES
( 1, 1, 1,'2020-11-18 07:10:25 +00:00'),
( 2, 2, 1,'2020-11-18 07:30:25 +00:00'),
( 3, 3, 1,'2020-11-18 07:50:25 +00:00'),
( 4, 2, 2,'2020-11-18 19:10:25 +00:00'),
( 5, 3, 2,'2020-11-18 19:22:38 +00:00'),
( 6, 1, 2,'2020-11-18 20:01:05 +00:00'),
( 7, 1, 1,'2020-11-19 07:11:23 +00:00'),
( 8, 1, 2,'2020-11-19 16:21:53 +00:00'), <-- Asia/Kuala_Lumpur +8 should be in 20.11 (refer to the check_outs field in the results output)
( 9, 1, 1,'2020-11-19 19:11:23 +00:00'), <-- Asia/Kuala_Lumpur +8 should be in 20.11 (refer to the check_ins field in the results output)
(10, 1, 2,'2020-11-19 20:21:53 +00:00'), <-- Asia/Kuala_Lumpur +8 should be in 20.11 (refer to the check_outs field in the results output)
(11, 1, 1,'2020-11-20 07:41:38 +00:00'),
(12, 1, 2,'2020-11-20 08:52:01 +00:00');
這是一個要測試的小提琴。
該查詢不包括時區 Asia/Kuala_Lumpur +8 的 output 中的第 8-10 行,盡管它應該。 結果顯示“行”字段11,12
。
如何修復查詢,以便它根據給定時區的日期生成報告? (意味着我可以將Asia/Kuala_Lumpur
更改為America/New_York
等)
我被告知要做這樣的事情:
where created_at >= timestamp '2020-11-20' AT TIME ZONE 'Asia/Kuala_Lumpur'
and created_at < timestamp '2020-11-20' AT TIME ZONE 'Asia/Kuala_Lumpur' + interval '1 day'
但我不確定如何應用它。 在這個小提琴中似乎無法正常工作。 它應該包括第 8、9、10、11、12 行,但只顯示第 8、9、10 行。
考慮對您的設置進行一些修改:
CREATE TABLE employee (
id int PRIMARY KEY -- !
, name text -- do NOT use char(n) !
, division_id int
);
CREATE TABLE attendance (
id int PRIMARY KEY --!
, employee_id int NOT NULL REFERENCES employee -- FK!
, activity_type int
, created_at timestamptz NOT NULL
);
定義 PK 可以更輕松地聚合行,因為 PK 覆蓋GROUP BY
子句中的整行。 看:
我不會使用“名稱”作為列名。 它不是描述性的。 每隔一列可以命名為“名稱”。 考慮:
SELECT *
FROM ( -- complete employee/date grid for division in range
SELECT g.d::date AS the_date, id AS employee_id, name, division_id
FROM (
SELECT generate_series(MIN(created_at) AT TIME ZONE 'Asia/Kuala_Lumpur'
, MAX(created_at) AT TIME ZONE 'Asia/Kuala_Lumpur'
, interval '1 day')
FROM attendance
) g(d)
CROSS JOIN employee e
WHERE e.division_id = 1
) de
LEFT JOIN ( -- checkins & checkouts per employee/date for division in range
SELECT employee_id, ts::date AS the_date
, array_agg(id) as rows
, min(ts) FILTER (WHERE activity_type = 1) AS min_check_in
, max(ts) FILTER (WHERE activity_type = 2) AS max_check_out
, array_agg(ts::time) FILTER (WHERE activity_type = 1) AS check_ins
, array_agg(ts::time) FILTER (WHERE activity_type = 2) AS check_outs
FROM (
SELECT a.id, a.employee_id, a.activity_type, a.created_at AT TIME ZONE 'Asia/Kuala_Lumpur' AS ts -- convert to timestamp
FROM employee e
JOIN attendance a ON a.employee_id = e.id
-- WHERE a.created_at >= timestamp '2020-11-20' AT TIME ZONE 'Asia/Kuala_Lumpur' -- "sargable" expressions
-- AND a.created_at < timestamp '2020-11-21' AT TIME ZONE 'Asia/Kuala_Lumpur' -- exclusive upper bound (includes all of 2020-11-20);
AND e.division_id = 1
ORDER BY a.employee_id, a.created_at, a.activity_type -- optional to guarantee sorted arrays
) sub
GROUP BY 1, 2
) a USING (the_date, employee_id)
ORDER BY 1, 2;
db<> 在這里擺弄
請注意,我的查詢輸出 Asia/Kuala_Lumpur 的本地日期和時間:
test=> SELECT timestamptz '2020-11-20 08:52:01 +0' AT TIME ZONE 'Asia/Kuala_Lumpur' AS local_ts;
local_ts
---------------------
2020-11-20 16:52:01
從哪兒開始? 您需要了解時區的概念和 Postgres 數據類型timestamp with time zone
( timestamptz
) 與timestamp without time zone
( timestamp
)。 否則,將是無休止的混亂。 從這里開始:
最值得注意的是, timestamptz
不存儲時區:
當簡單地將timestamptz
轉換為date
或timestamp
,假設 session 的當前時區設置。 不是你想要的。 使用AT TIME ZONE
構造顯式提供時區以避免這種問題。 在你的小提琴中,你有兩個:
...
, generate_series(
startdate::timestamp AT TIME ZONE 'Asia/Kuala_Lumpur',
enddate::timestamp AT TIME ZONE 'Asia/Kuala_Lumpur',
interval '1 day') g(d)
...
也沒有做你想做的事。 在(錯誤!)轉換為timestamp
之后, AT TIME ZONE
構造將值轉換回timestamptz
。
此外,您的查詢會生成所有用戶的完整笛卡爾積以及表attendance
的最大天數,只是為了將其減少回一天:
where created_at >= timestamp '2020-11-20' AT TIME ZONE 'Asia/Kuala_Lumpur'
and created_at < timestamp '2020-11-20' AT TIME ZONE 'Asia/Kuala_Lumpur' + interval '1 day'
WHERE
子句最終完成了它應該做的事情。 但是首先生成完整的天數,然后丟棄大部分天數是沒有意義的。 (似乎你同時從我的另一個小提琴中復制了它?)
我注釋掉了WHERE
子句,並在我的查詢中保留了您的generate_series()
的優化版本作為概念證明。 進一步閱讀:
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