[英]MySQL or MariaDB spread Date range in single row into multiple series of rows
Example tables are below,示例表如下,
CREATE TABLE TIMEOFF_INFO (
TIMEOFF_ID INT PRIMARY KEY,
TIMEOFF_BEGIN DATE NOT NULL,
TIMEOFF_END DATE NOT NULL
)
and example rows below,和下面的示例行,
+------------+---------------+-------------+
| timeoff_id | timeoff_begin | timeoff_end |
+------------+---------------+-------------+
| 1 | 2021-10-01 | 2021-10-02 |
+------------+---------------+-------------+
| 2 | 2021-11-15 | 2021-11-15 |
+------------+---------------+-------------+
| 3 | 2021-12-18 | 2021-12-20 |
+------------+---------------+-------------+
What I want to get is convert table above to below so that I can join using each date in range.我想要得到的是将上面的表格转换为下面的表格,以便我可以使用范围内的每个日期加入。
2021-10-01 (id: 1's begin date)
2021-10-02 (id: 1's end date)
2021-11-15 (id: 2's begin and end date)
2021-12-18 (id: 3's begin date)
2021-12-19
2021-12-20 (id: 3's end date)
Is there any way to extend date range in single row to series of Date row?有没有办法将单行中的日期范围扩展到一系列日期行?
You could use a union all on the same table您可以在同一张桌子上使用联合
select id, timeoff_begin timeoff, concat(id, ' begin date') msg
from TIMEOFF_INFO
union all
select id, timeoff_end, concat(id, ' end date')
from TIMEOFF_INFO
order by id, timeoff
WITH RECURSIVE
daterange AS ( SELECT MIN(TIMEOFF_BEGIN) dStart, MAX(TIMEOFF_END) dEnd
FROM TIMEOFF_INFO ),
calendar AS ( SELECT dStart `date`
FROM daterange
UNION ALL
SELECT `date` + INTERVAL 1 DAY
FROM calendar
WHERE `date` < ( SELECT dEnd FROM daterange ) )
SELECT calendar.`date`
FROM calendar
WHERE EXISTS ( SELECT NULL
FROM TIMEOFF_INFO
WHERE calendar.`date` BETWEEN TIMEOFF_INFO.TIMEOFF_BEGIN AND TIMEOFF_INFO.TIMEOFF_END )
MariaDB 10.2+ or MySQL 8+ needed.需要 MariaDB 10.2+ 或 MySQL 8+。
Much the same as @akina but less sophisticated.与@akina 大致相同,但不那么复杂。 Here a cte is used simply to identify those where the date difference is > 1 then join to a dates table and unioned.此处 cte 仅用于识别日期差大于 1 的那些,然后加入日期表并合并。 the begin, intermediate and end dates are identified by a flag which is used to order by.开始日期、中间日期和结束日期由用于排序的标志标识。 This could be done without a cte but a cte does help clarify(i think), Note I have amended the sample data slightly这可以在没有 cte 的情况下完成,但 cte 确实有助于澄清(我认为),注意我已经稍微修改了示例数据
drop table if exists t;
create table t
(timeoff_id int, timeoff_begin date, timeoff_end date);
insert into t values
( 1 , '2020-10-01' , '2020-10-05'),
( 2 , '2020-11-15' , '2020-11-15'),
( 3 , '2020-12-18' , '2020-12-21');
with cte as
(select 2 as beg,timeoff_id , timeoff_begin, timeoff_end ,datediff(timeoff_end, timeoff_begin) diff
from t
where datediff(timeoff_end, timeoff_begin) > 1)
select cte.beg,cte.timeoff_id,dates.dte
from cte
join dates where dates.dte > cte.timeoff_begin and dates.dte <= date_add(cte.timeoff_begin, interval cte.diff -1 day)
union all
(select 1 as beg,timeoff_id , timeoff_begin dt from t
union all
select 3 ,timeoff_id , timeoff_end from t
)
order by timeoff_id, beg,dte;
+-----+------------+------------+
| beg | timeoff_id | dte |
+-----+------------+------------+
| 1 | 1 | 2020-10-01 |
| 2 | 1 | 2020-10-02 |
| 2 | 1 | 2020-10-03 |
| 2 | 1 | 2020-10-04 |
| 3 | 1 | 2020-10-05 |
| 1 | 2 | 2020-11-15 |
| 3 | 2 | 2020-11-15 |
| 1 | 3 | 2020-12-18 |
| 2 | 3 | 2020-12-19 |
| 2 | 3 | 2020-12-20 |
| 3 | 3 | 2020-12-21 |
+-----+------------+------------+
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