SQL Exercise: Providers of primary care who are not department heads
SQL hospital Database: Exercise-39 with Solution
39. From the following table, write a SQL query to identify those patients whose primary care is provided by a physician who is not the head of any department. Return Patient name as "Patient", Physician Name as "Primary care Physician".
Sample table: patient
Sample table: department
Sample table: physician
SELECT pt.name AS "Patient", p.name AS "Primary care Physician" FROM patient pt JOIN physician p ON pt.pcp=p.employeeid WHERE pt.pcp NOT IN (SELECT head FROM department);
Patient | Primary care Physician -------------------+------------------------ John Smith | John Dorian Grace Ritchie | Elliot Reid Random J. Patient | Elliot Reid Dennis Doe | Christopher Turk (4 rows)
Explanation: The said query in SQL that selects the names of patients and their physicians who are not heads of any department.
The JOIN statement combines the 'patient' and 'physician' tables based on the pcp and employeeid columns.
The WHERE clause filters the results to ensure that only pcp's who are not department heads and this is accomplished using a subquery. The subquery that selects the "head" column from the 'department' table and filters out any pcp's who match this value using the NOT IN operator.
E R Diagram of Hospital Database:
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Previous SQL Exercise: Names of all patients who had at least 2 appointments.
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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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