SQueriosL
1.0.0
BUAT TABEL karyawan ( Employee_id INTEGER UNSIGNED BUKAN NULL AUTO_INCREMENT, nama belakang VARCHAR(30) BUKAN NULL, nama depan VARCHAR(30) BUKAN NULL, email VARCHAR(100) BUKAN NULL, tanggal_pekerjaan TANGGAL BUKAN NULL, catatan MEDIUMTEXT, PRIMARY KEY (employee_id), INDEX (nama_belakang), UNIK (email) );
MESIN=InnoDB;
BUAT alamat TABEL ( Employee_id INTEGER UNSIGNED NOT NULL, alamat VARCHAR(50) NOT NULL, kota VARCHAR(30) NOT NULL, negara bagian CHAR(2) NOT NULL, kode pos CHAR(5) NOT NULL, FOREIGN KEY (employee_id) REFERENSI pegawai ( karyawan_id) )
MESIN=InnoDB;
BUAT TABEL charset_example ( -> id INTEGER UNSIGNED BUKAN NULL AUTO_INCREMENT, -> ascii_string VARCHAR(255) CHARACTER SET ascii NOT NULL, -> latin1_string VARCHAR(255) CHARACTER SET latin1 BUKAN NULL, -> utf8_string VARCHAR(255) CHARACTER SET utf8 BUKAN NULL, -> KUNCI UTAMA (id) -> );
MASUKKAN KE karyawan(nama_depan, nama_belakang, email, tanggal_pekerjaan) NILAI ('Nischal', 'Bhatia', '[email protected]', '15-12-2014');
INSERT INTO Employee(nama_depan, nama_belakang, email, tanggal_pekerjaan) VALUES('Gurjot','Singh','[email protected]','2017-12-08');
INSERT INTO Employee(nama_depan, nama_belakang, email, tanggal_pekerjaan) VALUES('Jaskaran','Singh','[email protected]','14-04-2017');
MASUKKAN KE karyawan(nama_depan, nama_belakang, email, tanggal_pekerjaan) NILAI ('Anjandeep', 'Singh', '[email protected]', '30-11-2017');
PILIH * DARI karyawan;
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';
-- UPDATE pelanggan SET last_name = 'DAY-WEBB' WHERE customer_id = 245;
-- PILIH id_pelanggan, nama_depan, nama_belakang, pembaruan_terakhir DARI pelanggan WHERE nama_depan = '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> PILIH * -> DARI foo f KIRI LUAR GABUNG 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> PILIH * -> DARI foo f KANAN LUAR GABUNG 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> PILIH * -> DARI foo GABUNG 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> BUAT LIHAT Aktor_Penampilan AS -> PILIH a.nama_pertama, a.nama_akhir, JUMLAH(fa.film_id) SEBAGAI Jumlah_Penampilan -> DARI film_actor fa GABUNG aktor a -> ON a.actor_id = fa.actor_id -> KELOMPOK OLEH fa.actor_id ;
mysql> PILIH * -> DARI 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> BUAT LIHAT cust_address(first_name, last_name, cust_last_update, address, add_last_update) AS -> PILIH c.first_name, c.last_name, c.last_update, a.address, a.last_update -> DARI pelanggan c GABUNG alamat a -> ON a.address_id = c.address_id;
mysql> BUAT LIHAT active_customer AS -> PILIH 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> PILIH JUMLAH(rental_id) SEBAGAI Num_Rented, customer_id -> DARI rental -> GROUP BY customer_id -> ORDER BY Num_Rented DESC -> LIMIT 100;
Are there customers whose rental habits show they have a favorite actor?
Lihat tabel lain yang ditentukan dalam database sakila dan amati bagaimana tabel tersebut mengikuti 3NF.
PILIH a.actor_id, a.first_name, a.last_name, f.film_id, f.title DARI film_actor fa GABUNG aktor a ON fa.actor_id = a.actor_id GABUNG film f ON fa.film_id = f.film_id;
