SQL Exercise: Physicians, a medical procedure without certification
SQL hospital Database: Exercise-31 with Solution
31. From the following tables, write a SQL query to find all physicians who have performed a medical procedure but are not certified to do so. Return Physician name as "Physician".
Sample table: physician
Sample table: undergoes
Sample table: trained_in
SELECT name AS "Physician" FROM physician WHERE employeeid IN ( SELECT undergoes.physician FROM undergoes LEFT JOIN trained_In ON undergoes.physician=trained_in.physician AND undergoes.procedure=trained_in.treatment WHERE treatment IS NULL );
Physician ------------------ Christopher Turk (1 row)
The said query in SQL retrieves the name of physicians who have not been trained in a particular medical procedure by looking for records in the 'undergoes' table where a corresponding record in the 'trained_in' table with the same physician and procedure does not exist.
From the outer query filters the results by selecting only those records where the "employeeid" column is found in a subquery.
In the subquery the LEFT JOIN keyword joins the 'undergoes' and 'trained_in' tables based on the common columns physician, procedure, and treatment.
The WHERE clause in the subquery filters the results of the join to include only those records where "treatment" is NULL.
E R Diagram of Hospital Database:
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Previous SQL Exercise: Make a report of specified queries.
Next SQL Exercise: Doctors do the same procedure but are not certified.
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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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