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