有没有办法向您展示一个示例数据库?
您可以使用dbfiddle.uk 创建测试表和查询。
假设您有以下测试表(称为 ORIGINAL)和数据:
表格
create table original (
jddate number
, starttime timestamp
, endtime timestamp
, nr number
, terminal varchar2( 100 )
, dep varchar2( 100 )
, doc number
, typ number
, key1 number
, key2 number
) ;
插入
insert into original
select 118001
, trunc( sysdate ) + ( 6/24 ) + ( 30/(24*60 ) )
, trunc( sysdate ) + (86399/86400)
, 34000001, 'MM01X11', 'XX01', 1000800001, 1, 99000000, 99000000 from dual union all
-- duplicate
select 118001
, trunc( sysdate ) + ( 6/24 ) + ( 30/(24*60 ) )
, trunc( sysdate ) + (86399/86400)
, 34000001, 'MM01X11', 'XX01', 1000800001, 1, 99000000, 99111111 from dual union all
select 118001
, trunc( sysdate ) + ( 6/24 ) + ( 30/(24*60 ) )
, trunc( sysdate ) + (86399/86400)
, 34000001, 'MM01X11', 'XX01', 1000800001, 1, 99000000, 99111111 from dual union all
--
select 118001
, trunc( sysdate ) + ( 6/24 ) + ( 30/(24*60 ) )
, trunc( sysdate ) + (86399/86400)
, 34000001, 'MM01X11', 'XX01', 1000800001, 2, 99000000, 99000000 from dual union all
select 118001
, trunc( sysdate ) + ( 6/24 ) + ( 30/(24*60 ) )
, trunc( sysdate ) + (86399/86400)
, 34000001, 'MM01X11', 'XX01', 1000800001, 3, 99000000, 99000000 from dual union all
select 118001
, trunc( sysdate ) + ( 6/24 ) + ( 30/(24*60 ) )
, trunc( sysdate ) + (86399/86400)
, 34000001, 'MM01X11', 'XX01', 1000800001, 4, 99000000, 99000000 from dual ;
选择
-- select * from original;
JDDATE STARTTIME ENDTIME NR TERMINAL DEP DOC TYP KEY1 KEY2
118001 15-DEC-18 06.30.00.000000000 15-DEC-18 23.59.59.000000000 34000001 MM01X11 XX01 1000800001 1 99000000 99000000
118001 15-DEC-18 06.30.00.000000000 15-DEC-18 23.59.59.000000000 34000001 MM01X11 XX01 1000800001 1 99000000 99111111
118001 15-DEC-18 06.30.00.000000000 15-DEC-18 23.59.59.000000000 34000001 MM01X11 XX01 1000800001 1 99000000 99111111
118001 15-DEC-18 06.30.00.000000000 15-DEC-18 23.59.59.000000000 34000001 MM01X11 XX01 1000800001 2 99000000 99000000
118001 15-DEC-18 06.30.00.000000000 15-DEC-18 23.59.59.000000000 34000001 MM01X11 XX01 1000800001 3 99000000 99000000
118001 15-DEC-18 06.30.00.000000000 15-DEC-18 23.59.59.000000000 34000001 MM01X11 XX01 1000800001 4 99000000 99000000
要求
{1} 起初我需要“Typ” = 1 和 JDDate =>118000 的所有值
{2} 然后我需要基于 START 和 END 之间的差异/时间步长
在正确的 JDDate/ 格式中。不幸的是,这里有一些重复,
基于JDDate,START;结尾;终端:
{3} 至少我在 Key1 和
键 2。所以 Key1 每次都包含一个数字。 Key2 包含
一个 0 或一个数字,它也是 Key1 中的一个数字。如果在 Key1 中
和 Key2 是相同的数字,两行都应该删除。
示例查询 - 作为起点(WHERE 子句将需要更多工作......)
select distinct -- {2} remove duplicates
jddate
, endtime - starttime as interval_ -- {2}
, nr
, terminal
, dep
, doc
, typ
, key1
, key2
from original
where typ = 1 and jddate > 118000 -- {1}
and key1 <> key2 -- {3}
;
-- result
JDDATE INTERVAL_ NR TERMINAL DEP DOC TYP KEY1 KEY2
118001 +00 17:29:59.000000 34000001 MM01X11 XX01 1000800001 1 99000000 99111111
为了将 JDDATE 列中的值转换为 Oracle DATE,您可以使用您之前获得的 answer 的代码创建一个小函数。 (您不必这样做,但它会从您的 SELECT 中删除一些“混乱”)例如
-- https://stackoverflow.com/questions/53743601/sql-julien-date-cyyddd-to-date
/*
select date '1900-01-01'
+ floor(118001 / 1000) * interval '1' year
+ (mod(118001, 1000) - 1) * interval '1' day
from dual;
*/
-- this is far from perfect, needs range checking, exception handling etc
create or replace function cyyddd_to_date ( cyyddd number ) return date
is
begin
return
date '1900-01-01'
+ floor( cyyddd / 1000 ) * interval '1' year
+ ( mod( cyyddd, 1000 ) - 1 ) * interval '1' day
;
end;
/
-- quick test
select
cyyddd_to_date( 118001 ) date_
, to_char( cyyddd_to_date( 118001 ), 'YYYY-MM-DD' ) datetime_
from dual;
-- result
DATE_ DATETIME_
01-JAN-18 2018-01-01
最终查询
select distinct -- {2} remove duplicates
to_char( cyyddd_to_date( jddate ), 'YYYY-MM-DD' ) date_
, endtime - starttime interval_ -- {2}
, nr
, terminal
, dep
, doc
, typ
, key1
, key2
from original
where typ = 1 and jddate > 118000 -- {1}
and key1 <> key2 -- {3}
;
-- result
DATE_ INTERVAL_ NR TERMINAL DEP DOC TYP KEY1 KEY2
2018-01-01 +00 17:29:59.000000 34000001 MM01X11 XX01 1000800001 1 99000000 99111111
使用 Oracle 12c 和 Oracle 11g 测试,dbfiddle here。