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)


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:

E R Diagram: SQL Hospital Database.

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Previous SQL Exercise: Patients with their doctors and medications.
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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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