﻿ SQL: Six-character employee names with R in 3rd position

# SQL Exercise: Six-character employee names with R in 3rd position

## SQL employee Database: Exercise-106 with Solution

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106. From the following table, write a SQL query to find those employees whose name is six characters in length and the third character must be 'R'. Return complete information about the employees.

Sample table: employees

Sample Solution:

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

Sample Output:

``` emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------
69324 | MARKER   | CLERK    |      67832 | 1992-01-23 | 1400.00 |            |   1001
(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 any two characters, followed by the letter 'R' from the 'employees' table.

With the WHERE clause, the length() function will check if the emp_name has exactly 6 characters, and the LIKE operator will match any two characters following 'R'.

## Practice Online

Sample Database: employee

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Previous SQL Exercise: List the names of those employees starting with A.
Next SQL Exercise: Employee start with 'A' and end with 'N', length six.

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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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