﻿ SQL: Employee start with 'A' and end with 'N' , length six

# SQL Exercise: Employee start with 'A' and end with 'N', length six

## SQL employee Database: Exercise-107 with Solution

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107. From the following table, write a SQL query to find those employees whose name is six characters in length, starting with 'A' and ending with 'N'. Return number of employees.

Sample table: employees

Sample Solution:

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

Sample Output:

``` emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------
64989 | ADELYN   | SALESMAN |      66928 | 1991-02-20 | 1700.00 |     400.00 |   3001
(1 row)
```

Explanation:

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

With the WHERE clause, the length() function is used to verify that emp_name contains exactly six characters, and the LIKE operator is used to match names that begin with A and end with 'N'.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: Six-character employee names with R in 3rd position.
Next SQL Exercise: Employees joined with a as the 2nd character of a month.

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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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