SQL Exercise: List the employees whose ID not starting with digit 68
SQL employee Database: Exercise-111 with Solution
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111. From the following table, write a SQL query to find those employees whose ID not start with the digit 68. Return employee ID, employee ID using trim function.
Sample table: employees
Sample Solution:
SELECT emp_id,
trim(to_char(emp_id,'99999'))
FROM employees
WHERE trim(to_char(emp_id,'99999')) NOT LIKE '68%';
Sample Output:
emp_id | btrim --------+------- 66928 | 66928 67832 | 67832 65646 | 65646 67858 | 67858 69062 | 69062 63679 | 63679 64989 | 64989 65271 | 65271 66564 | 66564 69000 | 69000 69324 | 69324 (11 rows)
Explanation:
The said query in SQL that return all "emp_id" values with leading zeros, except those that start with "68".
The "trim" function removes any leading or trailing spaces from the resulting string.
The query filters out any rows where the "emp_id" starts with the characters "68" using the "NOT LIKE" operator.
Practice Online
Sample Database: employee

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Previous SQL Exercise: List the employees those who joined in 90's.
Next SQL Exercise: List the employees whose names containing the letter A.
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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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