SQL Exercise: Patients and their physicians who do not need a nurse
SQL hospital Database: Exercise-18 with Solution
18. From the following tables, write a SQL query to identify those patients and their physicians who do not require any nursing assistance. Return Name of the patient as "Name of the patient", Name of the Physician as "Name of the physician" and examination room as "Room No.".
Sample table: patient
Sample table: appointment
Sample table: physician
SELECT t.name AS "Name of the patient", p.name AS "Name of the physician", a.examinationroom AS "Room No." FROM patient t JOIN appointment a ON a.patient=t.ssn JOIN physician p ON a.physician=p.employeeid WHERE a.prepnurse IS NULL;
Name of the patient | Name of the physician | Room No. ---------------------+-----------------------+---------- John Smith | Christopher Turk | C Dennis Doe | Percival Cox | C (2 rows)
The said query in SQL that returns information about appointments where no preparatory nurse was assigned, including the patient's name, the physician's name, and the examination room number.
The JOIN keyword performs a join between the 'patient', 'appointment', and 'physician' tables based on their relational columns.
The 'appointment' and the 'patient' tables are joins based on the patient and ssn columns, and the 'appointment' and 'physician' tables are joins based on the the physician and employeeid columns.
The WHERE clause filters the results by appointments where no preparatory nurse was assigned. This is determined by checking whether the preparatory nurse identifier in the 'appointment' table is NULL.
E R Diagram of Hospital Database:
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Previous SQL Exercise: Patients with an appointment on the given date.
Next SQL Exercise: Names of patients, 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;
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