### 1. Find the addresses (location_id, street_address, city, state_province, country_name) of all the departments
SELECT location_id, street_address, city, state_province, country_name
FROM locations
JOIN departments ON locations.location_id = departments.location_id;
### 2. Find the name (first_name, last name), department ID and name of all the employees
SELECT first_name, last_name, department_id, department_name
FROM employees
JOIN departments ON employees.department_id = departments.department_id;
### 3. Find the name (first_name, last_name), job, department ID and name of the employees who works in London
SELECT first_name, last_name, job_title, department_id, department_name
FROM employees
JOIN departments ON employees.department_id = departments.department_id
WHERE city = 'London';
### 4. Write a query to find the employee id, name (last_name) along with their manager_id and name (last_name)
SELECT employee_id, last_name, manager_id, manager_last_name
FROM employees
JOIN employees AS m ON employees.manager_id = m.employee_id;
### 5. Write a query to find the name (first_name, last_name) and hire date of the employees who was hired after 'Jones'
SELECT first_name, last_name, hire_date
FROM employees
WHERE last_name > 'Jones';
### 6. Write a query to get the department name and number of employees in the department
SELECT department_name, COUNT(*) AS number_of_employees
FROM departments
JOIN employees ON departments.department_id = employees.department_id
GROUP BY department_name;
### 7. Write a query to display department name, name (first_name, last_name), hire date, salary of the manager for all managers whose experience is more than 15 years
SELECT department_name, first_name, last_name, hire_date, salary
FROM employees
JOIN departments ON employees.department_id = departments.department_id
WHERE manager_id IS NOT NULL AND
TIMESTAMPDIFF(YEAR, hire_date, CURRENT_DATE) > 15;
### 8. Write a query to find the name (first_name, last_name) and the salary of the employees who have a higher salary than the employee whose last_name='Bull'
SELECT first_name, last_name, salary
FROM employees
WHERE salary > (SELECT salary
FROM employees
WHERE last_name = 'Bull');
### 9. Write a query to find the name (first_name, last_name) of all employees who works in the IT department
SELECT first_name, last_name
FROM employees
JOIN departments ON employees.department_id = departments.department_id
WHERE department_name = 'IT';
### 10. Write a query to find the name (first_name, last_name) of the employees who have a manager and worked in a USA based department
SELECT first_name, last_name
FROM employees
JOIN departments ON employees.department_id = departments.department_id
JOIN locations ON departments.location_id = locations.location_id
WHERE manager_id IS NOT NULL AND country_name = 'USA';
### 11. Write a query to find the name (first_name, last_name), and salary of the employees whose salary is greater than the average salary
SELECT first_name, last_name, salary
FROM employees
WHERE salary > (SELECT AVG(salary)
FROM employees);
### 12. Write a query to find the name (first_name, last_name), and salary of the employees whose salary is equal to the minimum salary for their job grade
SELECT first_name, last_name, salary
FROM employees
JOIN jobs ON employees.job_id = jobs.job_id
WHERE salary = jobs.minimum_salary;
### 13. Write a query to find the name (first_name, last_name), and salary of the employees who earns more than the average salary and works in any of the IT departments
SELECT first_name, last_name, salary
FROM employees
JOIN departments ON employees.department_id = departments.department_id
WHERE salary > (SELECT AVG(salary)
FROM employees)
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MySQL is a open-source, free and very popular relational database management system which is developed, distributed and supported by Oracle corporation.
CREATE TABLE table_name (
column1 datatype,
column2 datatype,
....);
CREATE TABLE EMPLOYEE (
empId INTEGER PRIMARY KEY,
name TEXT NOT NULL,
dept TEXT NOT NULL
);
ALTER TABLE Table_name ADD column_name datatype;
INSERT INTO EMPLOYEE VALUES (0001, 'Dave', 'Sales');
TRUNCATE table table_name;
DROP TABLE table_name;
RENAME TABLE table_name1 to new_table_name1;
--Line1;
/* Line1,
Line2 */
INSERT INTO table_name (column1, column2, column3, ...) VALUES (value1, value2, value3, ...);
Note: Column names are optional.
INSERT INTO EMPLOYEE VALUES (0001, 'Ava', 'Sales');
SELECT column1, column2, ...
FROM table_name
[where condition];
SELECT * FROM EMPLOYEE where dept ='sales';
UPDATE table_name
SET column1 = value1, column2 = value2, ...
WHERE condition;
UPDATE EMPLOYEE SET dept = 'Sales' WHERE empId='0001';
DELETE FROM table_name where condition;
DELETE from EMPLOYEE where empId='0001';
CREATE INDEX index_name on table_name(column_name);
CREATE UNIQUE INDEX index_name on table_name(column_name);
DROP INDEX index_name ON table_name;
Creating a View:
CREATE VIEW View_name AS
Query;
SELECT * FROM View_name;
ALTER View View_name AS
Query;
DROP VIEW View_name;
CREATE TRIGGER trigger_name trigger_time trigger_event
ON tbl_name FOR EACH ROW [trigger_order] trigger_body
/* where
trigger_time: { BEFORE | AFTER }
trigger_event: { INSERT | UPDATE | DELETE }
trigger_order: { FOLLOWS | PRECEDES } */
DROP TRIGGER [IF EXISTS] trigger_name;
CREATE PROCEDURE sp_name(p1 datatype)
BEGIN
/*Stored procedure code*/
END;
CALL sp_name;
DROP PROCEDURE sp_name;
SELECT * FROM TABLE1 INNER JOIN TABLE2 where condition;
SELECT * FROM TABLE1 LEFT JOIN TABLE2 ON condition;
SELECT * FROM TABLE1 RIGHT JOIN TABLE2 ON condition;
SELECT select_list from TABLE1 CROSS JOIN TABLE2;