SQueriosL
1.0.0
СОЗДАТЬ ТАБЛИЦУ сотрудника ( идентификатор сотрудника INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, фамилия VARCHAR(30) NOT NULL, имя VARCHAR(30) NOT NULL, адрес электронной почты VARCHAR(100) NOT NULL, дата найма DATE NOT NULL, примечания MEDIUMTEXT, ПЕРВИЧНЫЙ КЛЮЧ (employee_id), INDEX (фамилия), УНИКАЛЬНЫЙ (электронная почта) );
ДВИГАТЕЛЬ = InnoDB;
СОЗДАТЬ ТАБЛИЦУ адрес (employer_id INTEGER UNSIGNED NOT NULL, адрес VARCHAR(50) NOT NULL, город VARCHAR(30) NOT NULL, состояние CHAR(2) NOT NULL, почтовый индекс CHAR(5) NOT NULL, ВНЕШНИЙ КЛЮЧ (employee_id) ССЫЛКИ на сотрудника ( идентификатор_сотрудника) )
ДВИГАТЕЛЬ = InnoDB;
CREATE TABLE charset_example ( -> id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, -> ascii_string VARCHAR(255) CHARACTER SET ascii NOT NULL, -> latin1_string VARCHAR(255) CHARACTER SET latin1 NOT NULL, -> utf8_string VARCHAR(255) НАБОР СИМВОЛЕЙ utf8 NOT NULL, -> ПЕРВИЧНЫЙ КЛЮЧ (id) -> );
INSERT INTO сотрудник(имя, фамилия, адрес электронной почты, дата найма) ЗНАЧЕНИЯ («Нишал», «Бхатия», «[email protected]», «15 декабря 2014 г.»);
INSERT INTO сотрудник(имя, фамилия, адрес электронной почты, дата найма) VALUES('Gurjot','Singh','[email protected]','2017-12-08');
INSERT INTO сотрудник(имя, фамилия, адрес электронной почты, дата найма) VALUES('Jaskaran','Singh','[email protected]','2017-04-14');
INSERT INTO сотрудник(имя, фамилия, адрес электронной почты, дата найма) ЗНАЧЕНИЯ («Анжандип», «Сингх», «[email protected]», «30 ноября 2017 г.»);
ВЫБРАТЬ * ОТ сотрудника;
INSERT INTO адрес(идентификатор_сотрудника,адрес,город,штат,почтовый индекс) VALUES(1, '123 Main Street', 'Anytowne', 'XE', '97052');
GROUP BY CLAUSE : :
- Suppose we want to find the top 10 customers who’ve spent the most money renting
movies.
SELECT customer_id, SUM(amount) AS Amt
FROM payment
-> GROUP BY customer_id
-> ORDER BY Amt DESC
-> LIMIT 10;
UPDATE CLAUSE : :
SELECT customer_id, first_name, last_name
FROM customer
WHERE first_name = 'Courtney' AND last_name = 'Day';
-- ОБНОВЛЕНИЕ клиента SET Last_name = 'DAY-WEBB' WHERE customer_id = 245;
-- ВЫБЕРИТЕ customer_id, first_name, Last_name, Last_update ОТ клиента WHERE first_name = 'Кортни';
DELETE CLAUSE : :
UPDATE customer
SET active = 0
WHERE customer_id = 245;
DELETE FROM rental / payment
WHERE customer_id = 245;
Now, we are allowed to delete the original P.K rows
DELETE FROM customer
WHERE customer_id = 245;
Retrieve all of the actors with “SON” in their last name and sort them alphabetically.
SELECT actor_id, last_name
FROM actor
WHERE last_name LIKE('%SON')
ORDER BY last_name ASC;
Calculate how many films there are for each rating category—G, PG, PG-13, R, and NC-17.
SELECT rating, COUNT(film_id) AS NumFilm
FROM film
GROUP BY rating;
What’s the ID of the customer who’s made the most visits to the video store?
SELECT customer_id, COUNT(payment_id) AS NumVisits, amount
FROM payment
GROUP BY customer_id
ORDER BY NumVisits DESC;
Let’s identify the top five actors who made the most film appearances and the
number of films they’ve each starred in.
SELECT actor_id, COUNT(actor_id) AS Appearances
FROM film_actor
GROUP BY actor_id
ORDER BY Appearances DESC
LIMIT 5;
Now, since actor_id is also present in the actor table
we use another SELECT statement to retrieve corresponding rows
SELECT actor_id, first_name, last_name
-> FROM actor
-> WHERE actor_id IN(107,102,198,181,23);
We have all the information we wanted, but unfortunately we still need to match
up the actors’ names and appearance counts manually
JOIN lets us do exactly that by connecting two or more tables based on a relationship that we specify.
SELECT a.first_name, a.last_name, COUNT(fa.actor_id) AS Appearance_Count
-> FROM film_actor fa JOIN actor a
-> ON a.actor_id = fa.actor_id
-> GROUP BY fa.actor_id
-> ORDER BY Appearance_Count DESC, a.first_name ASC, a.last_name ASC
-> LIMIT 5;
JOINS : : 3 Types : :
CREATE TABLE foo(foo_id INTEGER, foo_value CHAR(3));
CREATE TABLE bar(bar_id INTEGER, bar_value CHAR(3), foo_id INTEGER);
INSERT INTO foo(foo_id, foo_value)
-> VALUES(1, 'foo');
INSERT INTO foo(foo_id, foo_value)
-> VALUES(2, 'bar');
INSERT INTO bar(bar_id, bar_value, foo_id)
-> VALUES(1, 'baz', 2);
INSERT INTO bar(bar_id, bar_value, foo_id)
-> VALUES(2, 'qux', 3);
SHOW TABLES;
+---------------+
| Tables_in_foo |
+---------------+
| bar |
| foo |
+---------------+
SELECT *
-> FROM foo f INNER JOIN bar b
-> ON b.foo_id = f.foo_id;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 2 | bar | 1 | baz | 2 |
+--------+-----------+--------+-----------+--------+
mysql> SELECT * -> FROM foo f LEFT OUTER JOIN bar b -> ON b.foo_id = f.foo_id;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 1 | foo | NULL | NULL | NULL |
| 2 | bar | 1 | baz | 2 |
+--------+-----------+--------+-----------+--------+
mysql> SELECT * -> FROM foo f ПРАВОЕ ВНЕШНЕЕ СОЕДИНЕНИЕ bar b -> ON b.foo_id = f.foo_id;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 2 | bar | 1 | baz | 2 |
| NULL | NULL | 2 | qux | 3 |
+--------+-----------+--------+-----------+--------+
mysql> SELECT * -> FROM foo JOIN bar;
+--------+-----------+--------+-----------+--------+
| foo_id | foo_value | bar_id | bar_value | foo_id |
+--------+-----------+--------+-----------+--------+
| 1 | foo | 1 | baz | 2 |
| 2 | bar | 1 | baz | 2 |
| 1 | foo | 2 | qux | 3 |
| 2 | bar | 2 | qux | 3 |
+--------+-----------+--------+-----------+--------+
ABSTRACTING WITH VIEWS : :
mysql> СОЗДАТЬ ПРОСМОТР Actor_Appearance AS -> ВЫБРАТЬ a.first_name, a.last_name, COUNT(fa.film_id) AS Appearance_Count -> FROM Film_actor fa ПРИСОЕДИНЯЙТЕСЬ к актеру a -> ON a.actor_id = fa.actor_id -> ГРУППИРОВАТЬ ПО fa.actor_id ;
mysql> SELECT * -> FROM Actor_Appearance -> ORDER BY Appearance_Count DESC -> LIMIT 5;
+------------+-----------+------------------+
| first_name | last_name | Appearance_Count |
+------------+-----------+------------------+
| GINA | DEGENERES | 42 |
| WALTER | TORN | 41 |
| MARY | KEITEL | 40 |
| MATTHEW | CARREY | 39 |
| SANDRA | KILMER | 37 |
+------------+-----------+------------------+
### Duplication may arise, if ::
SELECT c.first_name, c.last_name, c.last_update, a.address, a.last_update
FROM customer AS c JOIN address AS a
ON a.address_id = c.address_id;
### Possible Solution is to provide alias or : :
mysql> CREATE VIEW cust_address(first_name, Last_name, cust_last_update, Address, add_last_update) AS -> ВЫБРАТЬ c.first_name, c.last_name, c.last_update, a.address, a.last_update -> ОТ клиента c JOIN адрес a -> ON a.address_id = c.address_id;
mysql> СОЗДАТЬ ПРОСМОТР active_customer AS -> ВЫБРАТЬ идентификатор клиента, имя, фамилия, адрес электронной почты -> ОТ клиента -> ГДЕ active = 1;
NORMALIZING FORMS : :
THIRD NF :
SELECT a.actor_id, a.first_name, a.last_name, f.film_id, f.title
FROM film_actor fa
JOIN actor a ON fa.actor_id = a.actor_id
JOIN film f ON fa.film_id = f.film_id;
ALTERING TABLES : :
mysql> ALTER TABLE actor
-> ADD COLUMN bio VARCHAR(255) AFTER last_name;
mysql> ALTER TABLE actor
-> DROP COLUMN bio;
mysql> ALTER TABLE actor
-> ADD INDEX idx_last_update(last_update);
mysql> ALTER TABLE actor
-> DROP INDEX idx_last_update;
EXERCISE : :
mysql> SELECT COUNT (rental_id) AS Num_Rented, customer_id -> ИЗ аренды -> ГРУППА ПО customer_id -> ORDER BY Num_Rented DESC -> LIMIT 100;
Are there customers whose rental habits show they have a favorite actor?
Просмотрите другие таблицы, определенные в базе данных sakila, и обратите внимание, как они соответствуют 3NF.
ВЫБЕРИТЕ a.actor_id, a.first_name, a.last_name, f.film_id, f.title ИЗ Film_actor fa ПРИСОЕДИНЯЙТЕСЬ к актеру a ON fa.actor_id = a.actor_id ПРИСОЕДИНЯЙТЕСЬ к фильму f ON fa.film_id = f.film_id;
