我有以下代码生成一张表,如图所示:
with test (code, datum) as
(select 600, date '2018-02-01' from dual union all
select 600, date '2018-02-02' from dual union all
select 0, date '2018-02-03' from dual union all
select 0, date '2018-02-04' from dual union all
select 0, date '2018-02-05' from dual union all
select 600, date '2018-02-06' from dual union all
select 600, date '2018-02-07' from dual union all
select 0, date '2018-02-08' from dual union all
select 0, date '2018-02-09' from dual
)
select * from test;
我尝试了以下操作,但未返回我需要的内容。
select * from (
select test.*, min(datum) over (partition by code order by code) as min_date,
max(datum) over (partition by code order by code) as max_date
from test) where min_date = datum;
我想实现的是仅在“代码”列上列出发生更改的记录(在发生更改的记录之前和之后)。
因此结果集应如下所示:
02/FEB/18 00:00:00 600
03/FEB/18 00:00:00 0
05/FEB/18 00:00:00 0
06/FEB/18 00:00:00 600
07/FEB/18 00:00:00 600
08/FEB/18 00:00:00 0
我提到了这个问题,但并没有解决我遇到的相同问题。
任何帮助表示感谢,谢谢。
更新:
这更接近我想要实现的目标。我可以列出列代码和更改不相同的所有行。但是,我需要在这些值也不同的地方列出记录。
with test (code, datum) as
(select 600, date '2018-02-01' from dual union all
select 600, date '2018-02-02' from dual union all
select 0, date '2018-02-03' from dual union all
select 0, date '2018-02-04' from dual union all
select 0, date '2018-02-05' from dual union all
select 600, date '2018-02-06' from dual union all
select 600, date '2018-02-07' from dual union all
select 0, date '2018-02-08' from dual union all
select 0, date '2018-02-09' from dual
)
,y1 as (
select test.datum, test.code, lead(code) over (order by datum) as change
from test
)
select * from y1;
最终结果集应仅包含突出显示的行。
更新2:
我想我可能做对了,仍然需要验证,但这似乎可行:
with test (code, datum) as
(select 600, date '2018-02-01' from dual union all
select 600, date '2018-02-02' from dual union all
select 0, date '2018-02-03' from dual union all
select 0, date '2018-02-04' from dual union all
select 0, date '2018-02-05' from dual union all
select 600, date '2018-02-06' from dual union all
select 600, date '2018-02-07' from dual union all
select 0, date '2018-02-08' from dual union all
select 0, date '2018-02-09' from dual
)
,y1 as (
select test.datum, test.code, lag(nvl(code,code)) over (order by datum) as after, lead(nvl(code,code)) over (order by datum) as before
from test
)
select * from y1 where code != before or code != after;
答案 0 :(得分:0)
不确定这是否有帮助,我看不出任何相关性来整理您的问题中的预期输出。
.m2
答案 1 :(得分:0)
以下脚本产生了预期的结果集:
with test (code, datum) as
(select 600, date '2018-02-01' from dual union all
select 600, date '2018-02-02' from dual union all
select 0, date '2018-02-03' from dual union all
select 0, date '2018-02-04' from dual union all
select 0, date '2018-02-05' from dual union all
select 600, date '2018-02-06' from dual union all
select 600, date '2018-02-07' from dual union all
select 0, date '2018-02-08' from dual union all
select 0, date '2018-02-09' from dual
)
,y1 as (
select test.datum, test.code, lag(nvl(code,code)) over (order by datum) as after, lead(nvl(code,code)) over (order by datum) as before
from test
)
select * from y1 where code != before or code != after;