我的表具有以下记录,分别记录不同的ID和不同的开始和结束日期
ID,开始日期,结束日期
1,2017-02-14,2018-11-05
我想使用日期维度表编写一个不包含的SQL,该表提供以下输出:基本上,从开始到结束日期之间每个月都有一条记录。
1, 2017, 02
1, 2017, 03
1, 2017, 04
1, 2017, 05
1, 2017, 06
1, 2017, 07
1, 2017, 08
1, 2017, 09
1, 2017, 10
1, 2017, 11
1, 2017, 12
1, 2018, 01
1, 2018, 02
1, 2018, 03
1, 2018, 04
1, 2018, 05
1, 2018, 06
1, 2018, 07
1, 2018, 09
1, 2018, 10
1, 2018, 11
答案 0 :(得分:1)
这就是我所做的,它像一种魅力:
-- sample data
WITH table_data
AS (
SELECT 1 AS id
,cast('2017-08-14' AS DATE) AS start_dt
,cast('2018-12-16' AS DATE) AS end_dt
UNION ALL
SELECT 2 AS id
,cast('2017-09-14' AS DATE) AS start_dt
,cast('2019-01-16' AS DATE) AS end_dt
)
-- find minimum date from the data
,starting_date (start_date)
AS (
SELECT min(start_dt)
FROM TABLE_DATA
)
--get all months between min and max dates
,all_dates
AS (
SELECT last_day(add_months(date_trunc('month', start_date), idx * 1)) month_date
FROM starting_date
CROSS JOIN _v_vector_idx
WHERE month_date <= add_months(start_date, abs(months_between((
SELECT min(start_dt) FROM TABLE_DATA), (SELECT max(end_dt) FROM TABLE_DATA))) + 1)
ORDER BY month_date
)
SELECT id
,extract(year FROM month_date)
,extract(month FROM month_date)
,td.start_dt
,td.end_dt
FROM table_data td
INNER JOIN all_dates ad
ON ad.month_date > td.start_dt
AND ad.month_date <= last_day(td.end_dt)
ORDER BY 1
,2
答案 1 :(得分:0)
您必须生成日期,然后从中选择年份和月份
select distinct year(date),month( date) from
(select * from (
select
date_add('2017-02-14 00:00:00.000', INTERVAL n5.num*10000+n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) as date
from
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n1,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n2,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n3,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n4,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n5
) a
where date >'2017-02-14 00:00:00.000' and date < '2018-11-05'
) as t
答案 2 :(得分:0)
请使用以下查询示例:
set @start_date = '2017-02-14';
set @end_date = LAST_DAY('2018-11-05');
WITH RECURSIVE date_range AS
(
select MONTH(@start_date) as month_, YEAR(@start_date) as year_, DATE_ADD(@start_date, INTERVAL 1 MONTH) as next_month_date
UNION
SELECT MONTH(dr.next_month_date) as month_, YEAR(dr.next_month_date) as year_, DATE_ADD(dr.next_month_date, INTERVAL 1 MONTH) as next_month_date
FROM date_range dr
where next_month_date <= @end_date
)
select month_, year_ from date_range
order by next_month_date desc