我有下表，为了便于阅读，我为每个案例编号分解了每位经理。
最后一列是第二列的累计总和。
理想情况下，月份的价值应该从1-3开始（就像你看到芝加哥的两个案例一样）。
但从表中可以看出，在某些情况下，某些条目缺失（标记为＆lt; -----）。
CITY CASE CASE_NUMBER MANAGER MONTH MONTHLY_TOTAL FISCAL_TOTAL
---------------------------------------------------------------------------
chicago case_1 1 John 1 2 2
chicago case_1 1 John 2 3 5
chicago case_1 1 John 3 5 10
chicago case_1 1 Jeff 1 4 4
chicago case_1 1 Jeff 2 2 6
chicago case_1 1 Jeff 3 3 9
chicago case_2 2 John 1 3 3
chicago case_2 2 John 2 2 5
chicago case_2 2 John 3 4 9
chicago case_2 2 Jeff 1 2 2
chicago case_2 2 Jeff 2 7 9 <----
newyork case_1 1 Lee 1 3 3
newyork case_1 1 Lee 2 4 7 <----
newyork case_1 1 Sue 1 2 2
newyork case_1 1 Sue 2 3 5
newyork case_1 1 Sue 3 2 7
newyork case_1 2 Lee 1 2 2
newyork case_1 2 Lee 2 4 6
newyork case_1 2 Lee 3 4 10
newyork case_1 2 Sue 1 3 3
newyork case_1 2 Sue 2 2 5 <----
我想要的是首先找出那些缺失的行并插入值。
对于那些缺少的人，month_total = 0
fiscal_total =上一行中的值。
E.g。对于第一个缺失的行应该是：
CITY CASE CASE_NUMBER MANAGER MONTH MONTHLY_TOTAL FISCAL_TOTAL
---------------------------------------------------------------------------
chicago case_2 2 Jeff 3 0 9
答案 0 :(得分：0)
也许这可能会提供一些想法。我在Postgresql上运行它，但只需要对Oracle进行微小的更改。然后一些清理。这个想法是这样的：
设定：
create table theTable(city varchar(64), theCase varchar(64), case_number int, manager varchar(64), month int, monthly_total int, fiscal_total int);
insert into theTable values('chicago', 'case_1', 1, 'John', 1, 2, 2);
insert into theTable values('chicago', 'case_1', 1, 'John', 2, 3, 5);
insert into theTable values('chicago', 'case_1', 1, 'John', 3, 5, 10);
insert into theTable values('chicago', 'case_1', 1, 'Jeff', 1, 4, 4);
insert into theTable values('chicago', 'case_1', 1, 'Jeff', 2, 2, 6);
insert into theTable values('chicago', 'case_1', 1, 'Jeff', 3, 3, 9);
insert into theTable values('chicago', 'case_2', 2, 'John', 1, 3, 3);
insert into theTable values('chicago', 'case_2', 2, 'John', 2, 2, 5);
insert into theTable values('chicago', 'case_2', 2, 'John', 3, 4, 9);
insert into theTable values('chicago', 'case_2', 2, 'Jeff', 1, 2, 2);
insert into theTable values('chicago', 'case_2', 2, 'Jeff', 2, 7, 9);
insert into theTable values('newyork', 'case_1', 1, 'Lee', 1, 3, 3);
insert into theTable values('newyork', 'case_1', 1, 'Lee', 2, 4, 7);
insert into theTable values('newyork', 'case_1', 1, 'Sue', 1, 2, 2);
insert into theTable values('newyork', 'case_1', 1, 'Sue', 2, 3, 5);
insert into theTable values('newyork', 'case_1', 1, 'Sue', 3, 2, 7);
insert into theTable values('newyork', 'case_1', 2, 'Lee', 1, 2, 2);
insert into theTable values('newyork', 'case_1', 2, 'Lee', 2, 4, 6);
insert into theTable values('newyork', 'case_1', 2, 'Lee', 3, 4, 10);
insert into theTable values('newyork', 'case_1', 2, 'Sue', 1, 3, 3);
insert into theTable values('newyork', 'case_1', 2, 'Sue', 2, 2, 5);
查询：
-- the 3 months
with three as (select 1 AS month union select 2 union select 3)
-- select rownum from all_tables where rownum < 4 -- Oracle world
-- the key
,city as (select distinct city, theCase, case_number, manager from theTable)
select x.city, x.theCase, x.case_number, x.manager, x.month, COALESCE(y.monthly_total,0)
,SUM(COALESCE(y.monthly_total,0)) OVER (PARTITION BY x.city, x.theCase, x.case_number, x.manager
ORDER BY x.month) AS fiscal_total2
,y.fiscal_total
-- cross product the key to form record for each month
from ( select c.city, c.theCase, c.case_number, c.manager, t.month
from city c
,three t
) x
-- join "desired" set back to "have" set
left outer join theTable y ON ( x.city = y.city
and x.theCase = y.theCase
and x.case_number = y.case_number
and x.manager = y.manager
and x.month = COALESCE(y.month, x.month)
)
order by x.city, x.theCase, x.case_number, x.manager, x.month;
结果：
city | thecase | case_number | manager | month | coalesce | fiscal_total2 |fiscal_total
---------+---------+-------------+---------+-------+----------+---------------+--------------
chicago | case_1 | 1 | Jeff | 1 | 4 | 4 | 4
chicago | case_1 | 1 | Jeff | 2 | 2 | 6 | 6
chicago | case_1 | 1 | Jeff | 3 | 3 | 9 | 9
chicago | case_1 | 1 | John | 1 | 2 | 2 | 2
chicago | case_1 | 1 | John | 2 | 3 | 5 | 5
chicago | case_1 | 1 | John | 3 | 5 | 10 | 10
chicago | case_2 | 2 | Jeff | 1 | 2 | 2 | 2
chicago | case_2 | 2 | Jeff | 2 | 7 | 9 | 9
chicago | case_2 | 2 | Jeff | 3 | 0 | 9 |
chicago | case_2 | 2 | John | 1 | 3 | 3 | 3
chicago | case_2 | 2 | John | 2 | 2 | 5 | 5
chicago | case_2 | 2 | John | 3 | 4 | 9 | 9
newyork | case_1 | 1 | Lee | 1 | 3 | 3 | 3
newyork | case_1 | 1 | Lee | 2 | 4 | 7 | 7
newyork | case_1 | 1 | Lee | 3 | 0 | 7 |
newyork | case_1 | 1 | Sue | 1 | 2 | 2 | 2
newyork | case_1 | 1 | Sue | 2 | 3 | 5 | 5
newyork | case_1 | 1 | Sue | 3 | 2 | 7 | 7
newyork | case_1 | 2 | Lee | 1 | 2 | 2 | 2
newyork | case_1 | 2 | Lee | 2 | 4 | 6 | 6
newyork | case_1 | 2 | Lee | 3 | 4 | 10 | 10
newyork | case_1 | 2 | Sue | 1 | 3 | 3 | 3
newyork | case_1 | 2 | Sue | 2 | 2 | 5 | 5
newyork | case_1 | 2 | Sue | 3 | 0 | 5 |
(24 rows)
答案 1 :(得分：0)
假设每个案例的每个经理都应该有三个条目，你可以尝试类似：
SELECT City, Case, Manager, COUNT(Manager) AS CASE_COUNT
FROM table(put actual table name)
GROUP BY City, Case, Manager
CASE_COUNT列将显示经理所处理的每个案例表格中的条目，如果该数字小于3，您知道缺少一个条目（当然这假设应该有每个人都有3个条目。）