我有专栏:
1。:delivery_date
列。它返回星期几,如下所示:
1 --monday
2 --tuesday
3 --wednesday
4 --thursday
5 --friday
2。:date
列。它返回取件的日期-这是正常日期列-例如:25.03.2019 (format DD.MM.YYYY)
但是,我们的系统会自动比较date
并将其设置为最接近的delivery_date
。
示例:
合作伙伴在周四和周二有delivery_date
每个星期 。[所以4和2]。他想在星期五提货。目前,我们的系统将date
更改为星期四。但是我想要下一个date
,即星期二-,我想将其解码为最新日期(格式DD.MM.YYYY)
输出如下:
--if pick up date: 22.03.2019 (friday)
--partner has delivery_date: 4,2(thursday, tuesday)
I want to get: 26.03.2019 --as output
我可以比较各列。..这没问题。.但是,我如何解码“实际提货”日期呢?
我该怎么做?
答案 0 :(得分:0)
对于解决此问题,构造“日历”表可能是一个优势-使其足够大(!),随后可用于压缩数据(请参见下面的代码中的注释)。为了进行测试,我们使用下表(Oracle 12c,18c):
表CAL _
create table cal_
as
select sysdate + level date_
, to_char( sysdate + level, 'Day' ) dname_
, to_char( sysdate + level, 'D' ) dnumber_
from dual
connect by level <= 31 ; -- number should be big enough to cover the lifetime of your system
SQL> select * from cal_ ;
DATE_ DNAME_ DNUMBER_
17-MAR-19 Sunday 7
18-MAR-19 Monday 1
19-MAR-19 Tuesday 2
20-MAR-19 Wednesday 3
21-MAR-19 Thursday 4
...
12-APR-19 Friday 5
13-APR-19 Saturday 6
14-APR-19 Sunday 7
15-APR-19 Monday 1
16-APR-19 Tuesday 2
表DELIVERYDAYS
create table deliverydays ( partner, deliveryday, dayofweek )
as
select
mod( level, 3 ) + 1
, mod( level, 5 ) + 1
, to_char( sysdate + ( mod( level, 5 ) + 2 ), 'Day' )
from dual
connect by level <= 10 ;
-- partner 4 delivers on Tue and Thu (same days as in the question)
insert into deliverydays ( partner, deliveryday, dayofweek )
values ( 4, 2, 'Tuesday' ) ;
insert into deliverydays ( partner, deliveryday, dayofweek )
values ( 4, 4, 'Thursday' ) ;
SQL> select * from deliverydays order by partner, deliveryday ;
PARTNER DELIVERYDAY DAYOFWEEK
1 2 Tuesday
1 4 Thursday
1 5 Friday
2 1 Monday
2 2 Tuesday
2 3 Wednesday
2 5 Friday
3 1 Monday
3 3 Wednesday
3 4 Thursday
4 2 Tuesday
4 4 Thursday
12 rows selected.
您可以DENSIFY这样来保存您的数据。请注意,交货日期之间的差距变得明显。 (带有下划线的列名称:来自CAL_表)。
select
partner, deliveryday , dayofweek
, date_, dname_, dnumber_
from deliverydays D partition by ( partner )
right join cal_ C on D.deliveryday = C.dnumber_
where partner = 4
order by date_ ;
PARTNER DELIVERYDAY DAYOFWEEK DATE_ DNAME_ DNUMBER_
4 NULL NULL 17-MAR-19 Sunday 7
4 NULL NULL 18-MAR-19 Monday 1
4 2 Tuesday 19-MAR-19 Tuesday 2
4 NULL NULL 20-MAR-19 Wednesday 3
4 4 Thursday 21-MAR-19 Thursday 4
4 NULL NULL 22-MAR-19 Friday 5
4 NULL NULL 23-MAR-19 Saturday 6
4 NULL NULL 24-MAR-19 Sunday 7
4 NULL NULL 25-MAR-19 Monday 1
4 2 Tuesday 26-MAR-19 Tuesday 2
4 NULL NULL 27-MAR-19 Wednesday 3
4 4 Thursday 28-MAR-19 Thursday 4
...
-- 31 rows selected
下一步,您可以使用LEAD()函数来查找“下一个”交货日期, 然后计算从“领取”到交货的天数,并根据需要设置日期格式。 (请参阅查询here的“注释”版本。)
select pid, dday, next_dday, date_, dname_, dnumber_
, case
when dnumber_ > next_dday then ( 7 - dnumber_ + next_dday ) -- next week
else next_dday - dnumber_
end ddiff_
, case
when dnumber_ > next_dday then
to_char( date_ + ( 7 - dnumber_ + next_dday ), 'DD.MM.YYYY' )
else
to_char( date_ + ( next_dday - dnumber_ ), 'DD.MM.YYYY' )
end delivered_on
from (
select
partner as pid, deliveryday as dday
, lead( deliveryday ) ignore nulls over ( order by date_ ) as next_dday
, dayofweek
, date_, dname_, dnumber_
from deliverydays D partition by ( partner )
right join cal_ C on D.deliveryday = C.dnumber_
where partner = 4 -- partner id
)
where to_char( date_, 'DD.MM.YYYY' ) = '22.03.2019' -- "pick up" date
order by date_ ;
结果
PID DDAY NEXT_DDAY DATE_ DNAME_ DNUMBER_ DDIFF_ DELIVERED_ON
4 NULL 2 22-MAR-19 Friday 5 4 26.03.2019
不使用外部WHERE子句,我们得到..
PID DDAY NEXT_DDAY DATE_ DNAME_ DNUMBER_ DDIFF_ DELIVERED_ON
4 NULL 2 17-MAR-19 Sunday 7 2 19.03.2019
4 NULL 2 18-MAR-19 Monday 1 1 19.03.2019
4 2 4 19-MAR-19 Tuesday 2 2 21.03.2019
4 NULL 4 20-MAR-19 Wednesday 3 1 21.03.2019
4 4 2 21-MAR-19 Thursday 4 5 26.03.2019
4 NULL 2 22-MAR-19 Friday 5 4 26.03.2019
4 NULL 2 23-MAR-19 Saturday 6 3 26.03.2019
4 NULL 2 24-MAR-19 Sunday 7 2 26.03.2019
4 NULL 2 25-MAR-19 Monday 1 1 26.03.2019
4 2 4 26-MAR-19 Tuesday 2 2 28.03.2019
4 NULL 4 27-MAR-19 Wednesday 3 1 28.03.2019
4 4 2 28-MAR-19 Thursday 4 5 02.04.2019
...
-- 31 rows selected