SQL Exercise: Find all physicians who completed a medical procedure
SQL hospital Database: Exercise-33 with Solution
33. From the following table, write a SQL query to find all physicians who completed a medical procedure with certification after the expiration date of their license. Return Physician Name as "Physician", Position as "Position".
Sample table: physician
Sample table: undergoes
Sample table: trained_in
SELECT name AS "Physician", position AS "Position" FROM physician WHERE employeeid IN ( SELECT physician FROM undergoes u WHERE date > ( SELECT certificationexpires FROM trained_in t WHERE t.physician = u.physician AND t.treatment = u.procedure ) );
Physician | Position --------------+------------------------------ Todd Quinlan | Surgical Attending Physician (1 row)
This SQL query retrieves the names and positions of physicians whose certification has expired for a procedure they have performed.
The WHERE clause filters the results where the "employeeid" column is found in a subquery.
The WHERE clause of the subquery include only those records where the "date" column is greater than the certification expiration date for the procedure.
The certification expiration date is obtained from inner most subquery which selects the "certificationexpires" column from the 'trained_in' table where the physician and procedure match those in the outer subquery.
E R Diagram of Hospital Database:
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Previous SQL Exercise: Doctors do the same procedure but are not certified.
Next SQL Exercise: Medical procedures done after certification expired.
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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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