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SQL Exercise: Find the floor with the most rooms available

SQL hospital Database: Exercise-26 with Solution

26. From the following tables, write a SQL query to find the floor where the maximum number of rooms are available. Return floor ID as "Floor", count "Number of available rooms".

Sample table: room


Sample Solution:

SELECT blockfloor as "Floor",
count(*) AS  "Number of available rooms"
FROM room
WHERE unavailable='false'
GROUP BY blockfloor
HAVING count(*) =
  (SELECT max(zz) AS highest_total
   FROM
( SELECTblockfloor ,
count(*) AS zz
      FROM room
      WHERE unavailable='false'
      GROUP BY blockfloor ) AS t );

Sample Output:

 Floor | Number of available rooms
-------+-----------------------
     1 |                     8
(1 row)

Explanation:

The given query in SQL that retrieves the floor and number of available rooms for each floor where rooms are available, groups the results by floor, and then returns only the floor(s) with the highest number of available rooms.

The WHERE clause filters the rows in the room table where the unavailable column is equal to 'false', indicating that the room is available.

The GROUP BY clause groups the rows by the blockfloor column, so that the count function is applied to each distinct floor.

The HAVING clause compares the count of available rooms on each floor to the highest_total value returned by the subquery. Only the floors with the same count as the highest_total are returned.

The subquery calculates the number of available rooms on each floor, and groups them by floor.

The MAX function in the outer query selects the highest number of available rooms across all floors.

Practice Online


E R Diagram of Hospital Database:

E R Diagram: SQL Hospital Database.

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Previous SQL Exercise: Count unoccupied rooms in each block and floor.
Next SQL Exercise: Find the floor with the minimum available rooms.

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SQL: Tips of the Day

Grouped LIMIT in PostgreSQL: Show the first N rows for each group?

db=# SELECT * FROM xxx;
 id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  3 |          1 | C
  4 |          1 | D
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
  8 |          2 | H
(8 rows)

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
(5 rows)

PostgreSQL v9.3 you can do a lateral join

select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
    select * from t t_inner
    where t_inner.section_id = t_outer.section_id
    order by t_inner.name
    limit 2
) t_top on true
order by t_outer.section_id;

Database: PostgreSQL

Ref: https://bit.ly/3AfYwZI

 

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