MySQL for business developer - Titouan BENOIT

MySQL for business developer
Titouan BENOIT – CTO at Welp.fr
titouan.benoit@gmx.fr

MySQL for business developer– Titouan BENOIT
Titouan BENOIT – CTO at Welp
– titouan@welp.today
– titouan.benoit@gmx.fr
MySQL for business developer - v1.0 2
Titouan BENOIT

Plan
1. Introduction
2. Query Language
3. Basics
– SELECT
– WHERE
– ORDER BY
– DISTINCT
– LIMIT & OFFSET
4. Advanced
– JOIN
– Sub-queries
– Math functions
– COUNT, SUM, AVG
5. phpMyAdmin
6. Examples/Exercices

MYSQL FOR BUSINESS DEVELOPER
1. Introduction
4

MySQL is an Open Source Relational Database Management System
(RDBMS)
– SQL (Structured Query Language) databases
– The second most used in the world
– Main purpose: storing data
In this presentation you will learn to:
– Make basics queries to search data into MySQL databases
– Extract data from MySQL databases
– Use PHPMyAdmin
– Compute statistics on data (count, average, …)
1. Introduction

1. Introduction: structure
MySQL Server
Databases
Welp OHA …
Tables
needs
propositions
users
…
id title author_id …
1 Mon titre1 5 …
2 Mon titre2 84 …
3 Mon titre3 6 …
… … … …
data
Columns
Rows

2. Query Language
7

SQL (Structured Query Language)
With SQL we can:
– Insert data
– Make queries (select data)
– Update data
– Delete data
– Create and modify schema (tables)
– Manage data access control
2. Query Language

3. Basics
9

SELECT * FROM database.table;
– Select all colums from table in the database
– Tips:
• USE database;
SELECT * FROM table
SELECT column1, column2, … FROM table
Alias: SELECT user.email FROM fos_user AS user
Example:
– SELECT id, email, first_name, last_name FROM welp_db.fos_user
3. Basics: SELECT

SELECT * FROM table WHERE condition;
– Select all colums from table in the database where condition is true
SELECT first_name, evaluation_average FROM fos_user WHERE evaluation_average = 5
– Select users where their average is equal to 5
Comparison operators Logical Operators
3. Basics: WHERE
Operator Description
= equal
< less than
<= less than or equal to
> greater than
>= greater than or equal to
<> or != different, not equal
<=> Equal (also to NULL value)
Operator Symbol Description
AND && AND
OR || OR
XOR Exclusive OR
NOT ! NOT
Example:

SELECT * FROM table ORDER BY column1;
– Select all colums from table and order by column1 descending order (default behavior)
SELECT * FROM table ORDER BY column1 DESC;
– Descending order
SELECT * FROM table ORDER BY column1 ASC;
– Ascending order
SELECT * FROM table ORDER BY column1, column2, …;
– Multicolumn order
– Note: the order of column after ORDER BY statement is important. The order will be done on the column1 and then column2, …
3. Basics: ORDER BY

SELECT DISTINCT column FROM table;
– Return only different rows (one by ‘column’ occurrences)
Example:
– SELECT DISTINCT species FROM animal;
– If I have 300 dogs, 400 cats and 250 elephants, this query will return:
• Dog
• Cat
• Elephant
– SELECT DISTINCT first_name FROM fos_user ORDER BY first_name ASC
3. Basics: DISTINCT

SELECT * FROM table LIMIT 0,10;
– Return 10 rows from the beginning (row number 0).
– How to use limit: LIMIT [offset, ]number_of_lines
– Note: the LIMIT statement must be at the end
Other syntaxe:
– LIMIT number_of_lines [OFFSET offset];
Examples
3. Basics: LIMIT & OFFSET

4. Advanced
15

INNER JOIN
– SELECT need.title, need.category_id, category.id, category.name FROM need
INNER JOIN category ON need.category_id = category.id
ORDER BY need.created_at
LIMIT 10;
– Note: inner join does not select rows where the relation is null (here where need has no category for example)
4. Advanced: JOIN
Need table Category table
need category
INNER JOIN

LEFT JOIN
– SELECT user.first_name, media.id AS mediaID, media.name AS avatar
FROM fos_user AS user
LEFT JOIN media ON user.avatar_id = media.id
ORDER BY user.created_at DESC
LIMIT 10, 10
4. Advanced: JOIN
User table Media table
media
LEFT JOIN

RIGHT JOIN
– SELECT category.id as categoryID, category.name AS categoryName, event.id AS eventID, event.title AS eventTitle
FROM event
RIGHT JOIN category ON event.category_id = category.id
4. Advanced: JOIN
Category table Event table
Difference with LEFT JOINThis category
has no event
This event has no category
RIGHT JOIN LEFT JOIN

Sub-queries
– In FROM statement:
• SELECT goodEvaluation.comment, goodEvaluation.evaluation, goodEvaluation.evaluator_id
FROM (
SELECT *
FROM evaluation
WHERE evaluation.evaluation = 5
) as goodEvaluation
WHERE goodEvaluation.evaluator_id = 19
– In WHERE clause:
• SELECT id, title, category_id
FROM need
WHERE category_id IN (
SELECT id
FROM category
WHERE name IN ('Actions à plusieurs', 'Visites de courtoisie')
);
4. Advanced: sub-queries
The sub-query:

Math functions
– CEIL(n) or CEILING(n): Return the smallest integer value not less than the argument
– FLOOR(n): Return the largest integer value not greater than the argument
– ROUND(n, d): Round the argument (d number of decimal)
– TRUNCATE(n, d): truncate the argument (d number of decimal)
– POW(n, e) or POWER(n, e): n^e
– SQRT(n): square root of the argument
– MOD(n, div): modulo of the division n/div
– RAND(): return a random value between 0 and 1
– SIGN(): return -1 for 0 and negative values and 1 for positive value
– ABS(): absolute value
– PI()
– COS()
– SIN()
4. Advanced: Math functions

Aggregate functions:
– COUNT(), SUM(), AVG(), MAX(), MIN()
SELECT COUNT(*) as nbNeed FROM need WHERE need.category_id = 1
SELECT AVG(fos_user.evaluation_average) as avgEvaluation FROM fos_user
GROUP BY & HAVING
– SELECT u.id, u.first_name, COUNT(n.author_id) need_count
FROM fos_user u
LEFT JOIN need n ON n.author_id = u.id
GROUP BY u.id
HAVING need_count > 0
ORDER BY need_count DESC
4. Advanced: COUNT, SUM, AVG

5. phpMyAdmin
22

Databases
5. phpMyAdmin

Tables
5. phpMyAdmin

Table structure
5. phpMyAdmin

Browse data
5. phpMyAdmin

SQL query
5. phpMyAdmin

Export data
5. phpMyAdmin

29

Requirements:
– Connect to Welp phpMyAdmin
– Login to prod with the read only user
1 slide with the problem at stake and the next slide the query
Good luck & have fun!

1°/ Retrieve the id, title and the
category_id of the 10 last created_at need

1°/ SOLUTION
SELECT id, title, category_id, created_at FROM need
ORDER BY created_at DESC LIMIT 10

2°/ Retrieve the id, first_name and the rank of fos_user
which has a rank more than or equal to 2 and less than 5
and sort by rank (descending)

2°/ SOLUTION
SELECT id, first_name, rank
FROM fos_user
WHERE rank >= 2 AND rank < 5
ORDER BY rank DESC

3°/ Retrieve all the distinct place_locality as city of the
need and sort them by alphabetical order

3°/ SOLUTION
SELECT DISTINCT(place_locality) as city
FROM need
ORDER BY city ASC

4°/ Retrieve the 12 last created need with their category.
Display the need id, title and the category id and name.

4°/ SOLUTION
SELECT need.id as needID, need.title as needTitle, need.category_id, category.id as
categoryID, category.name as categoryName
FROM need
INNER JOIN category ON need.category_id = category.id
ORDER BY need.created_at DESC
LIMIT 12

5°/ Retrieve fos_user id, first_name, last_name and for
each user display their number of welpActions.
Hint: use GROUP BY user.id

5°/ SOLUTION
SELECT user.id, user.first_name, user.last_name,
COUNT(welp_action.user_id) as WelpActions
FROM fos_user as user
LEFT JOIN welp_action ON user.id = welp_action.user_id
GROUP BY user.id
ORDER BY WelpActions DESC

MySQL for business developer
– https://en.wikipedia.org/wiki/MySQL
– https://en.wikipedia.org/wiki/SQL
– http://dev.mysql.com/
– https://openclassrooms.com/courses/administrez-vos-bases-de-donnees-avec-mysql
– http://dev.mysql.com/doc/refman/5.7/en/logical-operators.html
– http://dev.mysql.com/doc/refman/5.7/en/comparison-operators.html
– http://dev.mysql.com/doc/refman/5.7/en/join.html
– https://dev.mysql.com/doc/refman/5.0/en/group-by-handling.html
– http://dev.mysql.com/doc/refman/5.7/en/subqueries.html
– http://dev.mysql.com/doc/refman/5.7/en/mathematical-functions.html
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