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MySQL Cheat Sheet

MySQL Cheat Sheet

Help with SQL commands to interact with a MySQL database

MySQL Locations

  • Mac /usr/local/mysql/bin
  • Windows /Program Files/MySQL/MySQL version/bin
  • Xampp /xampp/mysql/bin

Add mysql to your PATH

# Current Session
export PATH=${PATH}:/usr/local/mysql/bin
# Permanantly
echo 'export PATH="/usr/local/mysql/bin:$PATH"' >> ~/.bash_profile

On Windows - https://www.qualitestgroup.com/resources/knowledge-center/how-to-guide/add-mysql-path-windows/

Login

mysql -u root -p

Show Users

SELECT User, Host FROM mysql.user;

Create User

CREATE USER 'someuser'@'localhost' IDENTIFIED BY 'somepassword';

Grant All Priveleges On All Databases

GRANT ALL PRIVILEGES ON * . * TO 'someuser'@'localhost';
FLUSH PRIVILEGES;

Show Grants

SHOW GRANTS FOR 'someuser'@'localhost';

Remove Grants

REVOKE ALL PRIVILEGES, GRANT OPTION FROM 'someuser'@'localhost';

Delete User

DROP USER 'someuser'@'localhost';

Exit

exit;

Show Databases

SHOW DATABASES

Create Database

CREATE DATABASE acme;

Delete Database

DROP DATABASE acme;

Select Database

USE acme;

Create Table

CREATE TABLE users(
id INT AUTO_INCREMENT,
   first_name VARCHAR(100),
   last_name VARCHAR(100),
   email VARCHAR(50),
   password VARCHAR(20),
   location VARCHAR(100),
   dept VARCHAR(100),
   is_admin TINYINT(1),
   register_date DATETIME,
   PRIMARY KEY(id)
);

Delete / Drop Table

DROP TABLE tablename;

Show Tables

SHOW TABLES;

Insert Row / Record

INSERT INTO users (first_name, last_name, email, password, location, dept, is_admin, register_date) values ('Brad', 'Traversy', 'brad@gmail.com', '123456','Massachusetts', 'development', 1, now());

Insert Multiple Rows

INSERT INTO users (first_name, last_name, email, password, location, dept,  is_admin, register_date) values ('Fred', 'Smith', 'fred@gmail.com', '123456', 'New York', 'design', 0, now()), ('Sara', 'Watson', 'sara@gmail.com', '123456', 'New York', 'design', 0, now()),('Will', 'Jackson', 'will@yahoo.com', '123456', 'Rhode Island', 'development', 1, now()),('Paula', 'Johnson', 'paula@yahoo.com', '123456', 'Massachusetts', 'sales', 0, now()),('Tom', 'Spears', 'tom@yahoo.com', '123456', 'Massachusetts', 'sales', 0, now());

Select

SELECT * FROM users;
SELECT first_name, last_name FROM users;

Where Clause

SELECT * FROM users WHERE location='Massachusetts';
SELECT * FROM users WHERE location='Massachusetts' AND dept='sales';
SELECT * FROM users WHERE is_admin = 1;
SELECT * FROM users WHERE is_admin > 0;

Delete Row

DELETE FROM users WHERE id = 6;

Update Row

UPDATE users SET email = 'freddy@gmail.com' WHERE id = 2;

Add New Column

ALTER TABLE users ADD age VARCHAR(3);

Modify Column

ALTER TABLE users MODIFY COLUMN age INT(3);

Order By (Sort)

SELECT * FROM users ORDER BY last_name ASC;
SELECT * FROM users ORDER BY last_name DESC;

Concatenate Columns

SELECT CONCAT(first_name, ' ', last_name) AS 'Name', dept FROM users;

Select Distinct Rows

SELECT DISTINCT location FROM users;

Between (Select Range)

SELECT * FROM users WHERE age BETWEEN 20 AND 25;

Like (Searching)

SELECT * FROM users WHERE dept LIKE 'd%';
SELECT * FROM users WHERE dept LIKE 'dev%';
SELECT * FROM users WHERE dept LIKE '%t';
SELECT * FROM users WHERE dept LIKE '%e%';

Not Like

SELECT * FROM users WHERE dept NOT LIKE 'd%';

IN

SELECT * FROM users WHERE dept IN ('design', 'sales');

Create & Remove Index

CREATE INDEX LIndex On users(location);
DROP INDEX LIndex ON users;

New Table With Foreign Key (Posts)

CREATE TABLE posts(
id INT AUTO_INCREMENT,
   user_id INT,
   title VARCHAR(100),
   body TEXT,
   publish_date DATETIME DEFAULT CURRENT_TIMESTAMP,
   PRIMARY KEY(id),
   FOREIGN KEY (user_id) REFERENCES users(id)
);

Add Data to Posts Table

INSERT INTO posts(user_id, title, body) VALUES (1, 'Post One', 'This is post one'),(3, 'Post Two', 'This is post two'),(1, 'Post Three', 'This is post three'),(2, 'Post Four', 'This is post four'),(5, 'Post Five', 'This is post five'),(4, 'Post Six', 'This is post six'),(2, 'Post Seven', 'This is post seven'),(1, 'Post Eight', 'This is post eight'),(3, 'Post Nine', 'This is post none'),(4, 'Post Ten', 'This is post ten');

INNER JOIN

SELECT
  users.first_name,
  users.last_name,
  posts.title,
  posts.publish_date
FROM users
INNER JOIN posts
ON users.id = posts.user_id
ORDER BY posts.title;

New Table With 2 Foriegn Keys

CREATE TABLE comments(
	id INT AUTO_INCREMENT,
    post_id INT,
    user_id INT,
    body TEXT,
    publish_date DATETIME DEFAULT CURRENT_TIMESTAMP,
    PRIMARY KEY(id),
    FOREIGN KEY(user_id) references users(id),
    FOREIGN KEY(post_id) references posts(id)
);

Add Data to Comments Table

INSERT INTO comments(post_id, user_id, body) VALUES (1, 3, 'This is comment one'),(2, 1, 'This is comment two'),(5, 3, 'This is comment three'),(2, 4, 'This is comment four'),(1, 2, 'This is comment five'),(3, 1, 'This is comment six'),(3, 2, 'This is comment six'),(5, 4, 'This is comment seven'),(2, 3, 'This is comment seven');

Left Join

SELECT
comments.body,
posts.title
FROM comments
LEFT JOIN posts ON posts.id = comments.post_id
ORDER BY posts.title;

Join Multiple Tables

SELECT
comments.body,
posts.title,
users.first_name,
users.last_name
FROM comments
INNER JOIN posts on posts.id = comments.post_id
INNER JOIN users on users.id = comments.user_id
ORDER BY posts.title;

Aggregate Functions

SELECT COUNT(id) FROM users;
SELECT MAX(age) FROM users;
SELECT MIN(age) FROM users;
SELECT SUM(age) FROM users;
SELECT UCASE(first_name), LCASE(last_name) FROM users;

Group By

SELECT age, COUNT(age) FROM users GROUP BY age;
SELECT age, COUNT(age) FROM users WHERE age > 20 GROUP BY age;
SELECT age, COUNT(age) FROM users GROUP BY age HAVING count(age) >=2;
@Zed-M

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Zed-M commented Apr 23, 2019

Thank you brad

@NobleKiwam

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NobleKiwam commented Apr 23, 2019

Just started going through. Thanks

@ridl27

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ridl27 commented Apr 23, 2019

Very useful! thx <3

@Niyokwizerwa250

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Niyokwizerwa250 commented Apr 24, 2019

You are awesome! Keep doing what you are doing for the industry

@saddamcrr7

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saddamcrr7 commented Apr 24, 2019

Thanks a lot brad

@fendiversace

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fendiversace commented Apr 25, 2019

Very useful!

@rohankewal

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rohankewal commented Apr 26, 2019

Very helpful! Thanks brad @bradtraversy

@sohel1999

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sohel1999 commented Apr 27, 2019

very useful!

@zhouyang159

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zhouyang159 commented Apr 27, 2019

xiexie ni de crash course :) (someone from China)

@YuanruiZhang

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YuanruiZhang commented Apr 29, 2019

Thanks alot Brad!

@Foss-Bee

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Foss-Bee commented May 4, 2019

Thanks very much @bradtraversy. You always make me love programming. Keep up with the good work.

@xelinel32

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xelinel32 commented May 6, 2019

thx, man)

@kailichou

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kailichou commented May 6, 2019

Hi, there
is there anyone also struggling to add mysql to the path?
Screenshot 2019-05-06 at 15 26 38

@LinmeiJ

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LinmeiJ commented May 20, 2019

THANK YOU SO MUCH brad! very helpful!

@mymmoonoa

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mymmoonoa commented Jun 9, 2019

nice work!

@abdalahshaban

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abdalahshaban commented Jun 12, 2019

nice work

@zaki1001

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zaki1001 commented Jun 12, 2019

thanks a ton mate

@ball97

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ball97 commented Jun 18, 2019

Thank you Brad. You are the best !!

@Ajmal0197

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Ajmal0197 commented Jun 18, 2019

Thanks master

@joshuaOgwang

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joshuaOgwang commented Jun 20, 2019

Thanks

@MbuguaCaleb

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MbuguaCaleb commented Jun 25, 2019

Thank you very Much Brad!!

@samdsg

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samdsg commented Jun 28, 2019

God Bless you Mr. Brad

@samdsg

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samdsg commented Jun 28, 2019

I have a question Mr. Brad.

$sql = 'SELECT c.categoryName as catname, m.img as img, p.id, p.product_name, p.product_desc FROM ' . $this->products . ' p RIGHT JOIN images m ON m.product_id = p.id LEFT JOIN categories c ON p.product_category = c.slug ORDER BY p.id DESC ';

i want to limit the image to only one image.. the image is a bucket with not less than four images.

I will really appreciate your response to me shortly.

@imaginaries

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imaginaries commented Jun 30, 2019

Very useful, thanks.

@matthewheimark

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matthewheimark commented Jul 9, 2019

Super useful, thank you!

@Zhixin-Jack-Wang

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Zhixin-Jack-Wang commented Jul 16, 2019

GOOD WORK, THANK YOU

@YosiLeibman

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YosiLeibman commented Jul 17, 2019

I have a question Mr. Brad.

$sql = 'SELECT c.categoryName as catname, m.img as img, p.id, p.product_name, p.product_desc FROM ' . $this->products . ' p RIGHT JOIN images m ON m.product_id = p.id LEFT JOIN categories c ON p.product_category = c.slug ORDER BY p.id DESC ';

i want to limit the image to only one image.. the image is a bucket with not less than four images.

I will really appreciate your response to me shortly.

there's a LIMIT clause. here you can read about.

@mlomelisa

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mlomelisa commented Aug 10, 2019

Very useful, thanks.

@sukhdevVyas

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sukhdevVyas commented Aug 16, 2019

It was a great refresher. Thank you :-) Expecting more videos or cheat seats for mySql administration.

@maggiesb

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maggiesb commented Sep 10, 2019

This is awesome! Thanks so much, Brad.

@delaumbriajohn

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delaumbriajohn commented Sep 12, 2019

Thank you for this , Sir!

@chrisvaughan83

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chrisvaughan83 commented Sep 23, 2019

Thanks! this is super helpful!

@lin3yx

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lin3yx commented Oct 20, 2019

Thank you so much for your wonderful work!

@domnet-webdesign

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domnet-webdesign commented Nov 7, 2019

Thanks you very much for this sample documentation

@mouadTaoussi

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mouadTaoussi commented Nov 7, 2019

Thank You Brad Traversy !
You're Work Apperciated Honestly!
Congratulations 900k Subscriber!

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