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

SQL employee Database: Exercise-105 with Solution

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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
----------
 ADELYN
 ADNRES
(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

employee database structure

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

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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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