SQL Exercise: List the employees along with department name
SQL employee Database: Exercise-52 with Solution
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52. From the following table, write a SQL query to find employees along with their department details. Return employee ID, employee name, job name, manager ID, hire date, salary, commission, department ID, and department name.
Sample table: employees
Sample table: department
Pictorial Presentation:

Sample Solution:
SELECT e.emp_id,
e.emp_name,
e.job_name,
e.manager_id,
e.hire_date,
e.salary,
e.commission,
e.dep_id,
d.dep_name
FROM employees e,
department d
WHERE e.dep_id = d.dep_id;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id | dep_name --------+----------+-----------+------------+------------+---------+------------+--------+---------- 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 | FINANCE 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 | MARKETING 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 | FINANCE 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 | AUDIT 67858 | SCARLET | ANALYST | 65646 | 1997-04-19 | 3100.00 | | 2001 | AUDIT 69062 | FRANK | ANALYST | 65646 | 1991-12-03 | 3100.00 | | 2001 | AUDIT 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 | AUDIT 64989 | ADELYN | SALESMAN | 66928 | 1991-02-20 | 1700.00 | 400.00 | 3001 | MARKETING 65271 | WADE | SALESMAN | 66928 | 1991-02-22 | 1350.00 | 600.00 | 3001 | MARKETING 66564 | MADDEN | SALESMAN | 66928 | 1991-09-28 | 1350.00 | 1500.00 | 3001 | MARKETING 68454 | TUCKER | SALESMAN | 66928 | 1991-09-08 | 1600.00 | 0.00 | 3001 | MARKETING 68736 | ADNRES | CLERK | 67858 | 1997-05-23 | 1200.00 | | 2001 | AUDIT 69000 | JULIUS | CLERK | 66928 | 1991-12-03 | 1050.00 | | 3001 | MARKETING 69324 | MARKER | CLERK | 67832 | 1992-01-23 | 1400.00 | | 1001 | FINANCE (14 rows)
Explanation:
The said statement in SQL that selects the employee ID, employee name, job name, manager ID, hire date, salary, commission, department ID, and department name for all employees in the employees table who belong to a department that exists in the department table.
An inner join happens to combine the data from the tables employees and department based on the dep_id column, which is common to both tables. This allows the query to match up each employee with their corresponding department based on the department ID.
The WHERE clause checks whether the dep_id column in the employees table matches the dep_id column in the department table.
Using aliases e and d can make the query easier to read and write by shortening the table names and reducing repetition.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: List all the employees who joined before or after 1991.
Next SQL Exercise: Employees earning 60000 or not working as analysts.
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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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