SQL Exercise: Count the number of available rooms in each block
SQL hospital Database: Exercise-22 with Solution
22. From the following table, write a SQL query to count the number of available rooms in each block. Sort the result-set on ID of the block. Return ID of the block as "Block", count number of available rooms as "Number of available rooms".
Sample table: room
SELECT blockcode AS "Block", count(*) "Number of available rooms" FROM room WHERE unavailable='false' GROUP BY blockcode ORDER BY blockcode;
Block | Number of available rooms -------+--------------------------- 1 | 9 2 | 10 3 | 10 (3 rows)
The said query in SQL that selects the block code and the number of available rooms for each block where the "unavailable" column in the "room" table is set to 'false'. The results are grouped by block code and ordered by block code.
The "count" function counts the number of rows in the "room" table where the "unavailable" column is set to 'false'. This count is aliased as "Number of available rooms".
The "WHERE" statement filters the results to only include rows where the "unavailable" column is set to 'false'.
The "GROUP BY" statement groups the results by block code.
The "ORDER BY" statement orders the results by block code in ascending order.
E R Diagram of Hospital Database:
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Previous SQL Exercise: Find patients without an appointment.
Next SQL Exercise: Count the number of available rooms in each floor.
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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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