﻿ SQL: List the names of those employees starting with A

# SQL Exercise: List the names of those employees starting with A

## SQL employee Database: Exercise-105 with Solution

[An editor is available at the bottom of the page to write and execute the scripts.]

105. From the following table, write a SQL query to identify those employees whose names begin with 'A' and are six characters long. Return employee name.

Sample table: employees

Sample Solution:

``````SELECT emp_name
FROM employees
WHERE emp_name LIKE 'A%'
AND length(emp_name)=6;
``````

Sample Output:

``` emp_name
----------
(2 rows)
```

Explanation:

The said query in SQL that selects the names of employees whose names start with 'A' and have a length of exactly 6 characters from the employees table.

The WHERE clause includes employees whose names starts with 'A' and the length() function to check if the length of the name is exactly 6 characters.

## Practice Online

Sample Database: employee

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous SQL Exercise: List the number of employees and average salary.
Next SQL Exercise: Six-character employee names with R in 3rd position.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.

﻿

## 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

We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook