CODEWITHSUNDEEP
sqloracle-databaseoracle12c
genericHCU
genericHCU

Reputation: 4446

Categorize Overlap Type - Oracle

I've been able to use How do I find the total number of used days in a month? to answer how many TOTAL days we've had cats and dogs which is helpful but I need to know how many days we had:

Cats only: 4
Dogs only: 5
Both: 6

Thank you in advance!

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'));

Upvotes: 4

Views: 74

Answers (2)

Lukasz Szozda
Lukasz Szozda

Reputation: 175954

Using conditional aggregation and inline calendar table:

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

Upvotes: 3

Gordon Linoff
Gordon Linoff

Reputation: 1270463

Oracle 12c supports recursive CTEs, so you can expand the data and then aggregate:

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;

This generates the result set as columns in a row, rather than separate rows.

Upvotes: 1

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