SQL Exercise: Display the unique department with jobs
SQL employee Database: Exercise-10 with Solution
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10. From the following table, write a SQL query to find the unique department with jobs. Return department ID, Job name.
Sample table: employees
SELECT DISTINCT dep_id, job_name FROM employees ;
dep_id | job_name --------+----------- 3001 | MANAGER 2001 | ANALYST 3001 | SALESMAN 1001 | MANAGER 1001 | PRESIDENT 2001 | MANAGER 2001 | CLERK 1001 | CLERK 3001 | CLERK (9 rows)
The said query in SQL that selects distinct combinations of the dep_id and job_name columns from the 'employees' table. The "DISTINCT" keyword ensures that each combination of dep_id and job_name appears only once in the result set, even if there are multiple employees with the same job and department.
A query like this might be used to get all unique job titles and departments in the company based on the "employees" table.
Relational Algebra Expression:
Relational Algebra Tree:
Sample Database: employee
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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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