﻿ SQL: Produce the output of employees

# SQL Exercise: Produce the output of employees

## SQL employee Database: Exercise-6 with Solution

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6. Write a query in SQL to produce the output of employees as follows:
Employee
JONAS(manager).

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT emp_name || '('|| lower(job_name)||')' AS "Employee"
FROM employees;
``````

Sample Output:

```      Employee
--------------------
KAYLING(president)
BLAZE(manager)
CLARE(manager)
JONAS(manager)
SCARLET(analyst)
FRANK(analyst)
SANDRINE(clerk)
TUCKER(salesman)
JULIUS(clerk)
MARKER(clerk)
(14 rows)
```

Explanation:

The said query in SQL that retrieves a result set that includes a list of concatenated strings, where each string represents an employee name and job name enclosed in parentheses and with the job name in lowercase.

The concatenation is performed using the concatenation operator || and a string literal '(' and ')', with the lower() function used to convert the "job_name" to lowercase and the AS keyword assigns the name "Employee" to the newly created column.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Produce the output of employees name and job name.
Next SQL Exercise: Employees with Hire date in format February 22, 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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