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SQL Exercise: Count unoccupied rooms in each block and floor

SQL hospital Database: Exercise-25 with Solution

25. From the following tables, write a SQL query to count the number of rooms that are unavailable in each block and on each floor. Sort the result-set on block floor, block code. Return the floor ID as "Floor", block ID as "Block", and number of unavailable as “Number of unavailable rooms".

Sample table: room


Sample Solution:

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

Sample Output:

 Floor | Block | Number of unavailable rooms
-------+-------+---------------------------
     1 |     2 |                         1
     2 |     1 |                         1
     2 |     2 |                         1
     3 |     1 |                         1
     3 |     3 |                         1
     4 |     1 |                         1
     4 |     3 |                         1
(7 rows)

Explanation:

The said query in SQL that selects the blockfloor and blockcode columns from the room table and counts the number of unavailable rooms.

The uses of WHERE clause only includes rows where the unavailable column has a value of 'true'.

The GROUP BY clause groups the results by blockfloor and blockcode, and the ORDER BY clause sorts the output by blockfloor and blockcode.

Practice Online


E R Diagram of Hospital Database:

E R Diagram: SQL Hospital Database.

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Previous SQL Exercise: Count the number of rooms in each block on each floor.
Next SQL Exercise: Find the floor with the most rooms available.

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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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