SQL Exercise: Count unique patients who came to examination room C
SQL hospital Database: Exercise-14 with Solution
14. From the following tables, write a SQL query to count the number of unique patients who have been scheduled for examination room 'C'. Return unique patients as "No. of patients got appointment for room C".
Sample table: appointment
Sample Solution:
SELECT count(DISTINCT patient) AS "No. of patients got appointment for room C"
FROM appointment
WHERE examinationroom='C';
Sample Output:
No. of patients got appointment for room C -------------------------------------------- 3 (1 row)
Explanation:
The said query in SQL that retrieves the number of unique patients who have had appointments in examination room C.
The WHERE statement ensures that only the appointments in examination room C are counted.
Practice Online
E R Diagram of Hospital Database:

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Previous SQL Exercise: Patients and doctors who gave them preliminary care.
Next SQL Exercise: Patients and the room number where they treated.
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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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