﻿ SQL exercises on employee Database: List the employees who joined in the month having char 'A' at any position - w3resource

# SQL exercises on employee Database: List the employees who joined in the month having char 'A' at any position

## SQL employee Database: Exercise-114 with Solution

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114. From the following table, write a SQL query to find those employees who joined in any month, but the month name contain the character ‘A’. Return complete information about the employees.

Sample table: employees

Sample Solution:

``````SELECT *
FROM employees
WHERE to_char (hire_date,'MONTH') LIKE '%A%';
``````

Sample Output:

``` emp_id | emp_name | job_name | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+----------+------------+------------+---------+------------+--------
66928 | BLAZE    | MANAGER  |      68319 | 1991-05-01 | 2750.00 |            |   3001
65646 | JONAS    | MANAGER  |      68319 | 1991-04-02 | 2957.00 |            |   2001
67858 | SCARLET  | ANALYST  |      65646 | 1997-04-19 | 3100.00 |            |   2001
64989 | ADELYN   | SALESMAN |      66928 | 1991-02-20 | 1700.00 |     400.00 |   3001
65271 | WADE     | SALESMAN |      66928 | 1991-02-22 | 1350.00 |     600.00 |   3001
68736 | ADNRES   | CLERK    |      67858 | 1997-05-23 | 1200.00 |            |   2001
69324 | MARKER   | CLERK    |      67832 | 1992-01-23 | 1400.00 |            |   1001
(7 rows)
```

## Practice Online

Sample Database: employee

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## SQL: Tips of the Day

MYSQL OR vs IN performance:

In many database servers, IN() is just a synonym for multiple OR clauses, because the two are logically equivalent. Not so in MySQL, which sorts the values in the IN() list and uses a fast binary search to see whether a value is in the list. This is O(Log n) in the size of the list, whereas an equivalent series of OR clauses is O(n) in the size of the list (i.e., much slower for large lists)

Ref: https://bit.ly/3PzLY69