SQueriosL
1.0.0
CREATE TABLE 従業員 (従業員 ID INTEGER UNSIGNED NOT NULL AUTO_INCREMENT、姓 VARCHAR(30) NOT NULL、名 VARCHAR(30) NOT NULL、電子メール VARCHAR(100) NOT NULL、採用日付 DATE NOT NULL、メモ MEDIUMTEXT、主キー (従業員 ID)、INDEX (姓)、一意 (電子メール) );
ENGINE=InnoDB;
CREATE TABLE address (employee_id INTEGER UNSIGNED NOT NULL、住所 VARCHAR(50) NOT NULL、都市 VARCHAR(30) NOT NULL、州 CHAR(2) NOT NULL、郵便番号 CHAR(5) NOT NULL、FOREIGN KEY (employee_id) REFERENCES 従業員 (従業員 ID) )
ENGINE=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) CHARACTER SET utf8 NOT NULL、 -> 主キー (id) -> );
INSERT INTO 従業員(名、姓、メールアドレス、採用日) VALUES ('Nischal', 'Bhatia', '[email protected]', '2014-12-15');
INSERT INTO 従業員(名、姓、電子メール、採用日) VALUES('Gurjot','Singh','[email protected]','2017-12-08');
INSERT INTO 従業員(名、姓、電子メール、採用日) VALUES('Jaskaran','Singh','[email protected]','2017-04-14');
INSERT INTO 従業員(名、姓、電子メール、採用日) VALUES ('Anjandeep', 'Singh', '[email protected]', '2017-11-30');
SELECT * FROM 従業員;
INSERT INTO address(従業員ID,住所,市区町村,州,郵便番号) 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';
-- UPDATE customer SET last_name = 'DAY-WEBB' WHERE customer_id = 245;
-- customer_id、first_name、last_name、last_update FROM customer WHERE first_name = 'Courtney'; を選択します。
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 RIGHT OUTER 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 |
| NULL | NULL | 2 | qux | 3 |
+--------+-----------+--------+-----------+--------+
mysql> SELECT * -> FROM foo JOIN バー;
+--------+-----------+--------+-----------+--------+
| 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> CREATE VIEW Actor_Appearance AS -> SELECT a.first_name、a.last_name、COUNT(fa.film_id) ASAppearance_Count -> FROM film_actor fa JOIN 俳優 a -> ON a.actor_id = fa.actor_id -> GROUP BY fa.actor_id ;
mysql> SELECT * -> FROM Actor_Appearance -> ORDER BYAppearance_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 -> SELECT c.first_name, c.last_name, c.last_update, a.address, a.last_update -> FROM customer c JOIN address a -> ON a.address_id = c.address_id;
mysql> CREATE VIEW active_customer AS -> SELECT customer_id、first_name、last_name、email -> FROM customer -> WHERE 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 -> FROM Rental -> GROUP BY customer_id -> ORDER BY Num_Rented DESC -> LIMIT 100;
Are there customers whose rental habits show they have a favorite actor?
sakila データベースに定義されている他のテーブルに目を通し、それらが 3NF にどのように従うかを観察してください。
SELECT a.actor_id, a.first_name, a.last_name, f.film_id, f.title FROM film_actor fa JOIN 俳優 a ON fa.actor_id = a.actor_id JOIN film f ON fa.film_id = f.film_id;
