Question 90205 in PostgreSQL Question posted in
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How to get rid of VIEW in this request – Postgresql

CREATE VIEW A1 AS
SELECT client_ID , COUNT(dog_id)
FROM test_clients
GROUP BY client_ID
HAVING COUNT(dog_id)=2;

CREATE VIEW A2 AS
SELECT filial , COUNT(A1.client_ID)
FROM A1
JOIN test_clients USING (client_ID)
GROUP BY filial
HAVING COUNT(A1.client_ID)>10;

SELECT COUNT(filial)
FROM A2;

As far as I understand, this can be done through a subquery, but how?

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2

Answers


  1. 0

    Burns down to:

    SELECT count(*)
FROM  (
   SELECT 1
   FROM  (
      SELECT client_id
      FROM   test_clients
      GROUP  BY 1
      HAVING count(dog_id) = 2
      ) a1
   JOIN   test_clients USING (client_id)
   GROUP  BY filial
   HAVING count(*) > 10
   ) a2;

    Assuming filial is defined NOT NULL.

    Probably faster to use a window function and get rid of the self-join:

    SELECT count(*)
FROM  (
   SELECT 1
   FROM  (
      SELECT filial
           , count(dog_id) OVER (PARTITION BY client_id) AS dog_ct
      FROM   test_clients
      ) a1
   WHERE  dog_ct = 2
   GROUP  BY filial
   HAVING count(*) > 10
   ) a2;

    Depending on your exact table definition we might be able to optimize a bit further …

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  2. 0

    A slight refractor of Erwin’s suggestion, just for you to play around with…

    The outer query works because…

    • the inner query happens first
    • the WHERE clause happens next
    • then the GROUP BY and HAVING clauses
    • then the SELECT clause (so the COUNT() OVER ())
    • finally the DISTINCT
    SELECT
  DISTINCT
  COUNT(filial) OVER ()
FROM
(
  SELECT
    filial,
    client_id,
    COUNT(dog_id) OVER (PARTITION BY client_id) AS client_dog_ct
  FROM
    test_clients
)
  count_dogs
WHERE
  client_dog_ct = 2
GROUP BY
  filial
HAVING
  COUNT(DISTINCT client_id) > 10

    You may or may not want the COUNT(DISTINCT client_id), its not clear. So, play with that too.

    I’m not saying it’s any better, just that it’s different and might help your learning.

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