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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