SQL Exercise: Count the number of unavailable rooms
SQL hospital Database: Exercise-7 with Solution
7. From the following table, write a SQL query to count the number of unavailable rooms. Return count as "Number of unavailable rooms".
Sample table: room
Sample Solution:
SELECT count(*) "Number of unavailable rooms"
FROM room
WHERE unavailable='true';
Sample Output:
Number of unavailable rooms --------------------------- 7 (1 row)
Explanation:
The said query in SQL that counts the number of unavailable rooms in the room table.
The query selects all columns (*) from the room table where the unavailable column is set to 'true'. The resulting output will be a single row with a single column, showing the number of unavailable rooms in the room table. The column will be labeled "Number of unavailable rooms".
Pictorial presentation:

Practice Online
E R Diagram of Hospital Database:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous SQL Exercise: Count the number available rooms.
Next SQL Exercise: Find the name and department of the physician.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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
- Weekly Trends
- Python Interview Questions and Answers: Comprehensive Guide
- Scala Exercises, Practice, Solution
- Kotlin Exercises practice with solution
- MongoDB Exercises, Practice, Solution
- SQL Exercises, Practice, Solution - JOINS
- Java Basic Programming Exercises
- SQL Subqueries
- Adventureworks Database Exercises
- C# Sharp Basic Exercises
- SQL COUNT() with distinct
- JavaScript String Exercises
- JavaScript HTML Form Validation
- Java Collection Exercises
- SQL COUNT() function
- SQL Inner Join
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