SQueriosL
1.0.0
CREATE TABLE 직원( 직원 ID INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, last_name VARCHAR(30) NOT NULL, first_name VARCHAR(30) NOT NULL, 이메일 VARCHAR(100) NOT NULL, 고용 날짜 DATE NOT NULL, 메모 MEDIUMTEXT, PRIMARY KEY(employee_id), INDEX (성), UNIQUE (이메일) );
엔진=InnoDB;
CREATE TABLE 주소 ( 직원 ID INTEGER UNSIGNED NOT NULL, 주소 VARCHAR(50) NOT NULL, 도시 VARCHAR(30) NOT NULL, 주 CHAR(2) NOT NULL, 우편번호 CHAR(5) NOT NULL, FOREIGN KEY (employee_id) REFERENCES 직원 ( 직원_ID) )
엔진=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, -> 기본 키(id) -> );
INSERT INTO 직원(이름, 성, 이메일, 고용_날짜) VALUES ('Nischal', 'Bhatia', '[email protected]', '2014-12-15');
INSERT INTO 직원(이름, 성, 이메일, 고용_날짜) VALUES('Gurjot','Singh','[email protected]','2017-12-08');
INSERT INTO 직원(이름, 성, 이메일, 고용_날짜) VALUES('Jaskaran','Singh','[email protected]','2017-04-14');
INSERT INTO 직원(이름, 성, 이메일, 고용_날짜) VALUES ('Anjandeep', 'Singh', '[email protected]', '2017-11-30');
SELECT * 직원으로부터;
INSERT INTO address(employee_id,address,city,state,postcode) 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';
-- 업데이트 고객 SET last_name = 'DAY-WEBB' WHERE customer_id = 245;
-- SELECT customer_id, first_name, last_name, last_update FROM 고객 WHERE first_name = 'Courtney';
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 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> CREATE VIEW Actor_Appearance AS -> SELECT a.first_name, a.last_name, COUNT(fa.film_id) AS Appearance_Count -> FROM film_actor fa JOIN actor a -> ON a.actor_id = fa.actor_id -> GROUP BY fa.actor_id ;
mysql> SELECT * -> Actor_Appearance에서 -> Appearance_Count DESC 기준으로 ORDER -> 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 고객 c JOIN 주소 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 rent -> GROUP BY customer_id -> ORDER BY Num_Rented DESC -> LIMIT 100;
Are there customers whose rental habits show they have a favorite actor?
sakila 데이터베이스에 정의된 다른 테이블을 살펴보고 이들이 3NF를 어떻게 따르는지 관찰하세요.
SELECT a.actor_id, a.first_name, a.last_name, f.film_id, f.title FROM film_actor fa JOIN 배우 a ON fa.actor_id = a.actor_id JOIN film f ON fa.film_id = f.film_id;
