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SQL Exercise: List the employee whose designation is ANALYST

SQL employee Database: Exercise-23 with Solution

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23. From the following table, write a SQL query to find the employee whose designation is ‘ANALYST’. Return employee name, job name and salary.

Sample table: employees


Pictorial Presentation:

SQL exercises on employee Database: List the name, job_name, and salary of any employee whose designation is ANALYST

Sample Solution:

SELECT emp_name,
       job_name,
       salary
FROM employees
WHERE job_name = 'ANALYST';

Sample Output:

 emp_name | job_name | salary
----------+----------+---------
 SCARLET  | ANALYST  | 3100.00
 FRANK    | ANALYST  | 3100.00
(2 rows)

Explanation:

The provided SQL statement retrieves the "emp_name", "job_name", and "salary" columns from the 'employees' table where the "job_name" column has a value of 'ANALYST'.

The WHERE clause filters the result set for those employees who are working in the designation 'ANALYST'.

Relational Algebra Expression:

Relational Algebra Expression: List the name, job_name, and salary of any employee whose designation is ANALYST.

Relational Algebra Tree:

Relational Algebra Tree: List the name, job_name, and salary of any employee whose designation is ANALYST.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: List the employees whose salaries are less than 3500.
Next SQL Exercise: List the employees who have joined in the year 1991.

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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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