分类重叠类型 - Oracle答案

【问题标题】：Categorize Overlap Type - Oracle分类重叠类型 - Oracle
【发布时间】：2019-07-15 15:27:07
【问题描述】：

我已经能够使用 How do I find the total number of used days in a month? 来回答我们养猫和狗的 TOTAL 天数，这很有帮助，但我需要知道我们养了多少天：

Cats only: 4
Dogs only: 5
Both: 6

提前谢谢你！

CREATE TABLE "ANIMALGUESTS" 
(   "ID" NUMBER, 
"GUESTNAME" VARCHAR2(20 BYTE), 
"GUESTTYPE" VARCHAR2(20 BYTE), 
"CHECKIN" DATE, 
"CHECKOUT" DATE
);


Insert into ANIMALGUESTS (ID,GUESTNAME,GUESTTYPE,CHECKIN,CHECKOUT) values (1,'Tom','Cat',to_date('01-JAN-19','DD-MON-RR'),to_date('10-JAN-19','DD-MON-RR'));
Insert into ANIMALGUESTS (ID,GUESTNAME,GUESTTYPE,CHECKIN,CHECKOUT) values (2,'Spike','Dog',to_date('03-JAN-19','DD-MON-RR'),to_date('05-JAN-19','DD-MON-RR'));
Insert into ANIMALGUESTS (ID,GUESTNAME,GUESTTYPE,CHECKIN,CHECKOUT) values (3,'Spike','Dog',to_date('08-JAN-19','DD-MON-RR'),to_date('12-JAN-19','DD-MON-RR'));
Insert into ANIMALGUESTS (ID,GUESTNAME,GUESTTYPE,CHECKIN,CHECKOUT) values (4,'Cherie','Cat',to_date('07-JAN-19','DD-MON-RR'),to_date('09-JAN-19','DD-MON-RR'));
Insert into ANIMALGUESTS (ID,GUESTNAME,GUESTTYPE,CHECKIN,CHECKOUT) values (5,'Tyke','Dog',to_date('10-JAN-19','DD-MON-RR'),to_date('15-JAN-19','DD-MON-RR'));

【问题讨论】：

第一个输出是你已经拥有的，还是你想通过一些查询生成的？
@TimBiegeleisen 抱歉，我不清楚，这就是我需要的结果。我已经能够使用链接的解决方案来回答cats: 10 and dogs: 11，而不管差距和重叠。

标签： sql oracle oracle12c

【解决方案1】：

使用条件聚合和内联日历表：

WITH cte AS (
  SELECT DATE '2019-01-01' + rownum -1 dt FROM DUAL CONNECT BY ROWNUM < 366
)
SELECT DISTINCT
  SUM(CASE WHEN COUNT(DISTINCT GUESTTYPE)=2 THEN 1 END) OVER() AS both,
  SUM(CASE WHEN COUNT(DISTINCT GUESTTYPE)=1 AND MIN(GUESTTYPE)='Cat' THEN 1 END) OVER() AS cats_only,
  SUM(CASE WHEN COUNT(DISTINCT GUESTTYPE)=1 AND MIN(GUESTTYPE)='Dog' THEN 1 END) OVER() AS dogs_only
FROM cte c
LEFT JOIN "ANIMALGUESTS" a ON c.dt BETWEEN a.CHECKIN AND a.CHECKOUT
GROUP BY dt;

db<>fiddle demo

【讨论】：

谢谢！这看起来会奏效。我必须在未来证明connect by，但我认为这会很好。

【解决方案2】：

Oracle 12c 支持递归 CTE，因此您可以扩展数据然后聚合：

with cte as (
      select checkin as dt, checkout, guesttype
      from ANIMALGUESTS
      union all
      select dt + 1, checkout, guesttype
      from cte
      where dt < checkout
     )
select sum(case when cats > 0 and dogs > 0 then 1 else 0 end) as both,
       sum(case when cats > 0 and dogs = 0 then 1 else 0 end) as cats_only,
       sum(case when cats = 0 and dogs > 0 then 1 else 0 end) as dogs_only
from (select dt, sum(case when guesttype = 'Cat' then 1 else 0 end) as cats,
             sum(case when guesttype = 'Dog' then 1 else 0 end) as dogs
      from cte
      group by dt
     ) cte;

这会将结果集生成为一行中的列，而不是单独的行。

【讨论】：