SQL Exercise: Find employees and departments with a given budget
SQL JOINS: Exercise-28 with Solution
From the following tables write a SQL query to find the departments with budgets more than Rs. 50000 and display the first name and last name of employees.
Sample table:emp_department
Sample table: emp_details
Sample Solution:
SELECT emp_details.emp_fname AS "First Name", emp_lname AS "Last Name"
FROM emp_details
INNER JOIN emp_department
ON emp_details.emp_dept = emp_department.dpt_code
AND emp_department.dpt_allotment > 50000;
Output of the Query:
First Name Last Name Alan Snappy Maria Foster Michale Robbin Enric Dosio Joseph Dosni Zanifer Emily Kuleswar Sitaraman Henrey Gabriel Alex Manuel George Mardy
Explanation:
The said SQL query is selecting specific columns from the emp_details table and renaming them as 'First Name' and 'Last Name'. It then joins the emp_details table with the emp_department table based on a matching value in the 'emp_dept' column of the emp_details table and the 'dpt_code' column of the emp_department table. The query also includes a filter that only returns the rows where the 'dpt_allotment' column in the emp_department table is greater than 50000.
The INNER JOIN clause is used which is used to only return the rows where there is a match in both tables, it discards the unmatched rows from both tables.
Practice Online
Query Visualization:
Duration:

Rows:

Cost:

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Previous SQL Exercise: Employee and sanction amount for their department.
Next SQL Exercise: Departments where more than two employees are working.
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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;
Database: PostgreSQL
Ref: https://bit.ly/3AfYwZI
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