mySQL questions

1. Using the employees database, select all columns from employees who were hired in October of any year. Limit the results to 50 rows. 	===>>> select * from employees where month(hire_date) = 10 limit 50;

2. Using the employees database, select all columns from employees database. Show only those employees who were hired in even-numbered days. Limit results to 100.
===>>> Select * from employees where day(hire_date) %2 = 0 limit 100;

3. (5 points) Display the last name and first name columns as one column labeled "Full Name" with the format "Last Name, First Name" and display the birth_date column labeled "Birth Date". Limit results to 50 rows.
	Output:
	Full Name			Birth Date
	Facello, Georgi		1953-09-02
	Simmel, Bezalel		1964-06-02
===>>> select comcat(last_name, ", ", first_name) as "Full Name", birth_date as "Birth Date" from employees limit 50


4. (10 points) Display all columns from the dept_emp table and display rows for the Marketing department. Limit to 50 rows.
	Output:
	emp_no			dept_no							from_date			to_date
	11833			(intentionally left blank)		1999-07-15			9999-01-01
	11841			(intentionally left blank)		1996-04-02			1996-10-24
===>>> select * from dept_emp where dept_no = 'd001' limit 50

5. (10 points) Show all the salaries for employees during the years 1990, 1993, and 1996. Limit to 100 rows.
	Output:
	emp_no		salary			from_date			to_date
	10011		56753	 		'1996-01-21'		'1996-11-09'
	10033		60433	 		'1993-03-16'		'1993-03-24'
	10082		48935	 		'1990-01-03'		'1990-01-15'
===>>> select * from salaries where year(from_date) in(1990, 1993, 1996) and year(to_date) in (1990, 1993, 1996) limit 100

6. (5 points) Count the number of male employees who were born in the year 1956. Label the column as "1956 Babies"
	Output:
	1956 Babies
	13822
===>>> select count(emp_no) as "1956 Babies" from employees where gender = 'm' and year(birth_date) = 1956

7. (10 points) For all employees with salaries less than 41,000. Display employee number labeled as "ID #", salary as "Salary" and add a column labeled "Bonus" where the value is 10% of the employee's salary. Limit to 100 rows.
	Output:
	ID #		Salary 			Bonus
	10003		40006			4000.6
	10004		40054			4005.4
	10006		40000			4000.0
===>>> select empno as "ID #", salary as "Salary", (salary * .1) as "Bonus" from salaries where salary < 41000 limit 100