表t1中有一些数据如下:
date dealer YTD_Value
2018-01 A 1100
2018-02 A 2000
2018-03 A 3000
2018-04 A 4200
2018-05 A 5000
2018-06 A 5500
2017-01 B 100
2017-02 B 200
2017-03 B 500
... ... ...
然后我想编写一个SQL查询此表并获得以下结果:
date dealer YTD_Value MTD_Value QTD_Value
2018-01 A 1100 1100 1100
2018-02 A 2000 900 2000
2018-03 A 3000 1000 3000
2018-04 A 4200 1200 1200
2018-05 A 5000 800 2000
2018-06 A 5500 500 2500
2017-01 B 100 100 100
2017-02 B 200 100 200
2017-03 B 550 350 550
... ... ... ... ...
'YTD'
表示迄今为止的年份
'MTD'
表示迄今为止的月份
'QTD'
表示迄今为止的季度
因此,如果我要计算MTD
中经销商QTD
的{{1}}和'A'
值,则该值应与'2018-01'
相同。
如果我要为YTD
中的经销商MTD
计算'A'
值,则'2018-06'
值应等于MTD
中的YTD
值减去'2018-06'
中的YTD
值。并且'2018-05'
中的QTD值应等于'2018-06'
中的YTD
值减去'2018-06'
中的YTD
值或等于( 2018-04,2018-05,2018-06)
对于B等其他经销商,同样的规则。
如何编写SQL来实现此目的?
答案 0 :(得分:2)
QTD计算很棘手,但是您可以在没有子查询的情况下执行此查询。基本思想是对月度值进行lag()
。然后使用max()
分析函数在季度开始时获取YTD值。
当然,一年的第一季度没有这样的值,因此需要coalesce()
。
尝试一下:
with t(dte, dealer, YTD_Value) as (
select '2018-01', 'A', 1100 from dual union all
select '2018-02', 'A', 2000 from dual union all
select '2018-03', 'A', 3000 from dual union all
select '2018-04', 'A', 4200 from dual union all
select '2018-05', 'A', 5000 from dual union all
select '2018-06', 'A', 5500 from dual union all
select '2017-01', 'B', 100 from dual union all
select '2017-02', 'B', 200 from dual union all
select '2017-03', 'B', 550 from dual
)
select t.*,
(YTD_Value - lag(YTD_Value, 1, 0) over (partition by substr(dte, 1, 4) order by dte)) as MTD_Value,
(YTD_Value -
coalesce(max(case when substr(dte, -2) in ('03', '06', '09') then YTD_VALUE end) over
(partition by substr(dte, 1, 4) order by dte rows between unbounded preceding and 1 preceding
), 0
)
) as QTD_Value
from t
order by 1
Here是db <>小提琴。
答案 1 :(得分:1)
以下查询应完成该任务。它使用CTE将varchar date列转换为日期,然后进行一些联接以恢复要比较的值。
我在this db fiddle中对其进行了测试,结果与您的预期结果相符。
WITH cte AS (
SELECT TO_DATE(my_date, 'YYYY-MM') my_date, dealer, ytd_value FROM my_table
)
SELECT
TO_CHAR(ytd.my_date, 'YYYY-MM') my_date,
ytd.ytd_value,
ytd.dealer,
ytd.ytd_value - NVL(mtd.ytd_value, 0) mtd_value,
ytd.ytd_value - NVL(qtd.ytd_value, 0) qtd_value
FROM
cte ytd
LEFT JOIN cte mtd ON mtd.my_date = ADD_MONTHS(ytd.my_date, -1) AND mtd.dealer = ytd.dealer
LEFT JOIN cte qtd ON qtd.my_date = ADD_MONTHS(TRUNC(ytd.my_date, 'Q'), -1) AND mtd.dealer = qtd.dealer
ORDER BY dealer, my_date
PS:date
是大多数RDBMS(包括Oracle)中的保留字,我在查询中将该列重命名为my_date
。
答案 2 :(得分:1)
您可以将lag()
Windows分析和sum() over ..
聚合函数用作:
select "date",dealer,YTD_Value,MTD_Value,
sum(MTD_Value) over (partition by qt order by "date")
as QTD_Value
from
(
with t("date",dealer,YTD_Value) as
(
select '2018-01','A',1100 from dual union all
select '2018-02','A',2000 from dual union all
select '2018-03','A',3000 from dual union all
select '2018-04','A',4200 from dual union all
select '2018-05','A',5000 from dual union all
select '2018-06','A',5500 from dual union all
select '2017-01','B', 100 from dual union all
select '2017-02','B', 200 from dual union all
select '2017-03','B', 550 from dual
)
select t.*,
t.YTD_Value - nvl(lag(t.YTD_Value)
over (partition by substr("date",1,4) order by substr("date",1,4) desc, "date"),0)
as MTD_Value,
substr("date",1,4)||to_char(to_date("date",'YYYY-MM'),'Q')
as qt,
substr("date",1,4) as year
from t
order by year desc, "date"
)
order by year desc, "date";