CODEWITHSUNDEEP
mysqlsqlmariadb
Gipyo.Choi
Gipyo.Choi

Reputation: 355

MySQL or MariaDB spread Date range in single row into multiple series of rows

Example tables are below,

CREATE TABLE TIMEOFF_INFO (
  TIMEOFF_ID INT PRIMARY KEY,
  TIMEOFF_BEGIN DATE NOT NULL,
  TIMEOFF_END DATE NOT NULL
)

and example rows below,

+------------+---------------+-------------+
| 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.

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?

Upvotes: 0

Views: 203

Answers (3)

P.Salmon
P.Salmon

Reputation: 17665

Much the same as @akina but less sophisticated. Here a cte is used simply to identify those where the date difference is > 1 then join to a dates table and unioned. the begin , intermediate and end dates are identified by a flag which is used to order by. This could be done without a cte but a cte does help clarify(i think), Note I have amended the sample data slightly

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 |
+-----+------------+------------+

Upvotes: 1

Akina
Akina

Reputation: 42844

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.

Upvotes: 1

ScaisEdge
ScaisEdge

Reputation: 133400

You could use a union all on the same table

    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

Upvotes: 0

Related Questions