w3resource

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


Sample Solution:

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

Sample Output:

 Block | Number of available rooms
-------+---------------------------
     1 |                         9
     2 |                        10
     3 |                        10
(3 rows)

Explanation:

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.

Practice Online


E R Diagram of Hospital Database:

E R Diagram: SQL Hospital Database.

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous SQL Exercise: Find patients without an appointment.
Next SQL Exercise: Count the number of available rooms in each floor.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.



Follow us on Facebook and Twitter for latest update.

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

 

We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook