我每周都有很多发货。我不希望他们都安排在星期五。我想在一周内将其拆分
例如如果我下周要寄出9批货物,我希望第一天2件,第二天2件,然后在本周剩余的时间每天1件。 我已经尝试了累加和,并且与其他数据(通过嵌套的for循环解决)有一个非常相似的问题,但是由于它在每周开始时一直在加权,这使我为此感到吃力。我试过了row_number并除以得到的天数,这会分割天数,但不会循环运动
-- Creates table 1
(SELECT 'SHIP1' SHIP_ID, '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP2', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP3', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP4', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP5', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP6', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP7', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP8', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP9', '202014' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP10', '202015' SHIP_WK FROM DUAL UNION ALL
SELECT 'SHIP11', '202015' SHIP_WK FROM DUAL) SHIPS
-- Creates table 2
(SELECT '202014' WEEK, '7' NO_DAYS FROM DUAL UNION ALL
SELECT '202015' WEEK, '5' NO_DAYS FROM DUAL) WEEK
我需要说SHIP1和SHIP2在202014年第1天进行,SHIP3 / 4在第2天进行,然后在第二天进行。 SHIP10 / 11从202015年的第1天和第2天开始,然后是3天(发货信息为空)。
答案 0 :(得分:4)
NTILE()
分析功能正是您所需要的。
with
ships (ship_id, ship_wk) as (
select 'SHIP1' , '202014' from dual union all
select 'SHIP2' , '202014' from dual union all
select 'SHIP3' , '202014' from dual union all
select 'SHIP4' , '202014' from dual union all
select 'SHIP5' , '202014' from dual union all
select 'SHIP6' , '202014' from dual union all
select 'SHIP7' , '202014' from dual union all
select 'SHIP8' , '202014' from dual union all
select 'SHIP9' , '202014' from dual union all
select 'SHIP10', '202015' from dual union all
select 'SHIP11', '202015' from dual)
, weeks (week, no_days) as (
select '202014', 7 from dual union all
select '202015', 5 from dual
)
select s.ship_id, s.ship_wk,
ntile(w.no_days) over (partition by s.ship_wk, w.no_days
order by s.ship_id) as day_no
from ships s left outer join weeks w on s.ship_wk = w.week
;
输出
SHIP_ID SHIP_W DAY_NO
--------- ------ ------
SHIP1 202014 1
SHIP2 202014 1
SHIP3 202014 2
SHIP4 202014 2
SHIP5 202014 3
SHIP6 202014 4
SHIP7 202014 5
SHIP8 202014 6
SHIP9 202014 7
SHIP10 202015 1
SHIP11 202015 2
这是从“船”端开始看,并分配一周中的几天。在您的原始帖子中,您还谈论一周中几天未使用的NULL;可以添加,但是不清楚从哪一侧看输出。您是否需要知道每个星期的每一天都有哪些船,还是需要为每艘船知道它们需要在哪一周的哪几周运送?