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
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;
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.
E R Diagram of Hospital Database:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous SQL Exercise: Patients with their doctors and medications.
Next SQL Exercise: Count the number of available rooms in each block.
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;
- 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
- 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