208 lines
7.2 KiB
SQL
208 lines
7.2 KiB
SQL
USE company;
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SELECT * FROM country;
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SELECT first_name "Firstname", last_name "Lastname" from employee;
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select distinct department_ID from employee;
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select count(job_id) from employee;
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select count(distinct job_id) from employee;
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-- liste des employés embauchés avant le 01.01.1991
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select * from employee where hire_date < '1991-01-01';
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-- FOnctions
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-- -- mettre en majuscule
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select upper(first_name) "Firstname" from employee;
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-- -- mettre les 3 premières lettres des nom et prénom des employés
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select substring(upper(first_name), 1, 3), substring(upper(last_name), 1, 3)
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from employee;
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-- -- prénom + nom + longueur totale du prénom+nom
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select first_name, last_name,
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char_length(first_name) + char_length(last_name) "Longueur totale"
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from employee;
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-- -- Qui gagne le plus dans l'entreprise
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select first_name, last_name, salary
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from employee order by salary desc limit 1;
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select first_name, last_name,
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salary >= (select max(salary) from employee)
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from employee order by salary desc;
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-- -- salaire min et max
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select min(salary), max(salary), avg(salary), sum(salary) from employee;
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-- -- les gens de l'IT qui gagnent moins de 5000
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select first_name, last_name, department_id, salary
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from employee where department_id = 60 and salary > 5000;
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-- -- les gens dans la vente (80) et le mercatique (20)
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select first_name, last_name, salary
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from employee where department_id in (20, 80);
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-- -- salaire entre 5000 et 10000
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select first_name, last_name, salary
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from employee where salary between 5000 and 10000;
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-- -- avec une commission
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select first_name, last_name, commission_pct
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from employee where commission_pct != 0;
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-- -- employés dont le nom de famille commance par "K"
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select first_name, last_name
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from employee where last_name like "K%";
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-- -- employés dont le prénom contient "an"
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select first_name, last_name
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from employee where first_name like "%an%";
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-- -- employés dont le prénom fait 4 lettres
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select first_name, last_name
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from employee where first_name like "____";
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-- -- REGEX : nom de famille commence par "S" ou "T"
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select first_name, last_name
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from employee where last_name regexp "^[ST]";
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-- -- le nombre de jour depuis l'embauche
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select first_name, last_name, hire_date,
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datediff(curdate(), hire_date) as nb_jours
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from employee order by nb_jours;
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-- Exercice
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-- -- les commerciaux gagnant plus de 9000 triés par commission décroissante
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select first_name, last_name, salary, commission_pct
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from employee
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where department_id = 80 and salary > 9000 order by commission_pct desc;
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-- -- les postes dont le salaire est au moins le double du salaire minimum
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select first_name, last_name, salary
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from employee where salary > (select min(salary) from employee) * 2
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order by salary asc;
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-- -- les employés dont le prénom commence et fini par la même lettre
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select first_name, last_name
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from employee
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where left(first_name, 1) = right(first_name, 1);
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-- -- le top 5 des employés les mieux payés n'étant pas chefs
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select first_name, last_name, salary, job_id
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from employee
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where employee_id not in (select manager_id from employee)
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order by salary desc limit 10;
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-- COURS
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-- -- Jointures
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-- -- -- https://cartman34.fr/informatique/sgbd/differences-entre-inner-left-right-et-outer-join-en-sql.html
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select first_name, last_name, department_name
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from employee e
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inner join department d
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on e.department_id = d.department_id;
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-- -- -- LEFT avec condition : les employés qui ne sont affectés à aucun départment
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select e.first_name, e.last_name, d.department_name
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from employee e
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left join department d
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on e.department_id = d.department_id
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where d.department_id is null;
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-- -- -- RIGHT avec condition : la liste des départments ne contenant aucun employé
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select e.first_name, e.last_name, d.department_name
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from employee e
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right join department d
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on e.department_id = d.department_id
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where e.department_id is null;
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-- -- -- FULL OUTER JOIN
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select e.first_name, e.last_name, d.department_name
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from employee e
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left join department d
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on e.department_id = d.department_id
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union
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select e.first_name, e.last_name, d.department_name
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from employee e
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right join department d
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on e.department_id = d.department_id;
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-- -- -- NATURAL JOIN -> INNER JOIN
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select location_id, street_address, city, country_name
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from location
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natural join country;
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-- -- -- EXEMPLE : nous voulons la liste des employés travaillant à Seattle
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-- -- -- On doit faire EMPLOYEE -> DEPARTMENT -> LOCATION
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select e.first_name, e.last_name, l.city
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from employee e
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join department d
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on e.department_id = d.department_id
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join location l
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on d.location_id = l.location_id
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where lower(l.city) = "seattle";
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-- -- -- EXEMPLE : nous voulons la liste des employés travaillant aux USA
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-- -- -- On doit faire EMPLOYEE -> DEPARTMENT -> LOCATION -> COUNTRY
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select e.first_name, e.last_name, l.city
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from employee e
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join department d
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on e.department_id = d.department_id
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join location l
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on d.location_id = l.location_id
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join country c
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on l.country_id = c.country_id
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where lower(c.country_name) like "%united%states%";
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-- -- -- Prénom et nom de l'employé suivi du prénom de son chef
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select e.first_name, e.last_name, m.first_name "Prénom chef", m.last_name "Nom chef"
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from employee e
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left join employee m
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on m.employee_id = e.manager_id;
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-- -- -- Nombre de gens dans chaque poste
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select j.job_title, count(*)
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from employee e
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join job j
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on e.job_id = j.job_id
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group by j.job_id;
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-- -- -- la liste des chefs et le nombre d'employés sous leurs ordres
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-- -- -- et dont le nombre d'employés est d'au moins 5
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select m.first_name "Prénom chef",
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m.last_name "Nom chef",
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count(e.employee_id) as count
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from employee e
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join employee m
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on m.employee_id = e.manager_id
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group by e.manager_id
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having count >= 5
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order by count desc;
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-- -- Sous-requêtes
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-- -- -- Tous les employés gagnant plus que l'employé nommé "Bull"
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select first_name, last_name, salary
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from employee
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where salary >
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(select salary from employee where lower(last_name) = "Bull");
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-- -- -- Tous les employés ayant pour chef un prénommé "Payam"
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-- -- -- JOINTURE
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select e.first_name, e.last_name, m.first_name "Prénom chef", m.last_name "Nom chef"
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from employee e
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left join employee m
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on m.employee_id = e.manager_id
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where lower(m.first_name) = "payam";
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-- -- -- Tous les employés ayant pour chef un prénommé "Payam"
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-- -- -- SOUS-REQUETE
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select first_name, last_name
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from employee
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where manager_id =
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(select employee_id from employee where lower(first_name) = "payam");
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-- -- -- Les employés dont le salaire est compris entre le plus petit et 3000
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select first_name, last_name, salary
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from employee
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where salary
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between (select min(salary) from employee) and 3000
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order by salary asc;
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-- -- -- les employés dont le chef bosse aux USA
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select first_name, last_name
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from employee
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where manager_id in
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(
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select employee_id from employee where department_id in
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(
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select department_id from department where location_id in
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(
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select location_id from location where lower(country_id) = "us"
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)
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)
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);
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-- -- Création de "Vues"
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-- -- Grosso modo, c'est une requête pré-enregistrée
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create view `seattle_employees` as
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select e.first_name, e.last_name, e.salary, d.department_name, l.city
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from employee e
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join department d
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on e.department_id = d.department_id
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join location l
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on d.location_id = l.location_id
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where lower(l.city) = "seattle";
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select * from seattle_employees; |