SQL Exercise: Find patients without an appointment
SQL hospital Database: Exercise-21 with Solution
21. From the following tables, write a SQL query to find those patients who did not schedule an appointment. Return Patient name as "Patient", Physician name as "Physician" and Medication name as "Medication".
Sample table: patient
Sample table: prescribes
Sample table: physician
Sample table: medication
Sample Solution:
SELECT t.name AS "Patient",
p.name AS "Physician",
m.name AS "Medication"
FROM patient t
JOIN prescribes s ON s.patient=t.ssn
JOIN physician p ON s.physician=p.employeeid
JOIN medication m ON s.medication=m.code
WHERE s.appointment IS NULL;
Sample Output:
Patient | Physician | Medication ------------+-------------+------------ Dennis Doe | Molly Clock | Thesisin (1 row)
Explanation:
The said query in SQL that selects the names of patients, physicians, and medications for all prescriptions where the appointment is null.
The "JOIN" statement connects the tables 'patient' and 'prescribes' based on the columns ssn and patient, the tables 'prescribes' and 'physician' based on the physician and employeeid columns, and the tables 'prescribes' and 'medication' are based on the columns medication and code.
The "WHERE" clause filters the results to only include prescriptions where the appointment is null.
Practice Online
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;
Database: PostgreSQL
Ref: https://bit.ly/3AfYwZI
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