SQL Exercise: Count the number of rooms in each block on each floor
SQL hospital Database: Exercise-24 with Solution
24. From the following tables, write a SQL query to count the number of available rooms for each floor in each block. Sort the result-set on floor ID, ID of the block. Return the floor ID as "Floor", ID of the block as "Block", and number of available rooms as "Number of available rooms".
Sample table: room
Sample Solution:
SELECT blockfloor AS "Floor",
blockcode AS "Block",
count(*) "Number of available rooms"
FROM room
WHERE unavailable='false'
GROUP BY blockfloor,
blockcode
ORDER BY blockfloor,
blockcode;
Sample Output:
Floor | Block | Number of available rooms -------+-------+--------------------------- 1 | 1 | 3 1 | 2 | 2 1 | 3 | 3 2 | 1 | 2 2 | 2 | 2 2 | 3 | 3 3 | 1 | 2 3 | 2 | 3 3 | 3 | 2 4 | 1 | 2 4 | 2 | 3 4 | 3 | 2 (12 rows)
Explanation:
The said query in SQL that selects the blockfloor and blockcode columns and counts the number of available rooms by grouping the data by blockfloor and blockcode where the value of unavailable column is 'false'.
The WHERE clause filters the data by only selecting rows where the value of the unavailable column is 'false'.
The GROUP BY clause groups the data by blockfloor and blockcode, which allows the count of available rooms to be aggregated by these columns.
The ORDER BY clause orders the results by blockfloor and blockcode in ascending order.
Practice Online
E R Diagram of 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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