SQL Exercise: Count the number of available rooms in each floor

SQL hospital Database: Exercise-23 with Solution

23. From the following table, write a SQL query to count the number of available rooms in each floor. Sort the result-set on block floor. Return floor ID as "Floor" and count the number of available rooms as "Number of available rooms".

Sample table: room

Sample Solution:

SELECT blockfloor AS "Floor",
       count(*) "Number of available rooms"
FROM room
WHERE unavailable='false'
GROUP BY blockfloor
ORDER BY blockfloor;

Sample Output:

 Floor | Number of available rooms
     1 |                         8
     2 |                         7
     3 |                         7
     4 |                         7
(4 rows)


The said query in SQL that retrieves the number of available rooms on each floor of a block in a building.

This query counts the number of rows that satisfy the WHERE condition, and the WHERE clause filters to include only rows where the unavailable column is 'false'.

The GROUP BY clause groups the data by the blockfloor column, so that the count of available rooms is computed for each distinct floor.

The ORDER BY clause sorts the output by the blockfloor column in ascending order.

Practice Online

E R Diagram of Hospital Database:

E R Diagram: SQL Hospital Database.

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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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