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:

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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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