테이블 조인
SQL Server 입문
John MacKintosh
Instructor
관계형 데이터베이스
기본 키
- 기본 키: 테이블의 각 행을 고유하게 식별
+-----------+-------------------+
| artist_id | name |
|-----------+-------------------|
| 1 | AC/DC |
| 2 | Accept |
| 3 | Aerosmith |
| 4 | Alanis Morissette |
| 5 | Alice In Chains |
+-----------+-------------------+
- 기본 키:
artist_id
+----------+-------------------------+-----------+
| album_id | title | artist_id |
|----------+-------------------------+-----------|
| 1 | For Those About To Rock | 1 |
| 2 | Balls to the Wall | 2 |
| 3 | Restless and Wild | 2 |
| 4 | Let There Be Rock | 1 |
| 5 | Big Ones | 3 |
+----------+-------------------------+-----------+
- 기본 키:
album_id artist_id는 무엇인가요?
외래 키
artist테이블
+-----------+-------------------+
| artist_id | name |
|-----------+-------------------|
| 1 | AC/DC |
| 2 | Accept |
| 3 | Aerosmith |
| 4 | Alanis Morissette |
| 5 | Alice In Chains |
+-----------+-------------------+
album테이블
+----------+-------------------------+-----------+
| album_id | title | artist_id |
|----------+-------------------------+-----------|
| 1 | For Those About To Rock | 1 |
| 2 | Balls to the Wall | 2 |
| 3 | Restless and Wild | 2 |
| 4 | Let There Be Rock | 1 |
| 5 | Big Ones | 3 |
+----------+-------------------------+-----------+
artist_id:artist에 대한 외래 키
album과 artist 조인
artist테이블
+-----------+-------------------+
| artist_id | name |
|-----------+-------------------|
| 1 | AC/DC |
| 2 | Accept |
| 3 | Aerosmith |
| 4 | Alanis Morissette |
| 5 | Alice In Chains |
+-----------+-------------------+
- AC/DC의
artist_id= 1
album테이블
+----------+-------------------------+-----------+
| album_id | title | artist_id |
|----------+-------------------------+-----------|
| 1 | For Those About To Rock | 1 |
| 2 | Balls to the Wall | 2 |
| 3 | Restless and Wild | 2 |
| 4 | Let There Be Rock | 1 |
| 5 | Big Ones | 3 |
+----------+-------------------------+-----------+
- 1행과 4행의
artist_id= 1
album과 artist 조인
+----------+-------------------------+-----------+-------------+
| album_id | title | artist_id | artist_name |
|----------+---------------------------------------+-----------|
| 1 | For Those About To Rock | 1 | AC/DC |
| 4 | Let There Be Rock | 1 | AC/DC |
+----------+-------------------------+-----------+-------------|
album테이블에서 앨범 정보 반환artist테이블에서 해당 아티스트 정보 반환artist_id열을 사용하여 조인
INNER JOIN
SELECT
album_id,
title,
album.artist_id,
name AS artist_name
FROM album
INNER JOIN artist ON artist.artist_id = album.artist_id
WHERE album.artist_id = 1;
+----------+-------------------------+-----------+-------------+
| album_id | title | artist_id | artist_name |
|----------+---------------------------------------+-----------|
| 1 | For Those About To Rock | 1 | AC/DC |
| 4 | Let There Be Rock | 1 | AC/DC |
+----------+-------------------------+-----------+-------------|
INNER JOIN 구문
SELECT
table_A.columnX,
table_A.columnY,
table_B.columnZ
FROM table_A
INNER JOIN table_B ON table_A.foreign_key = table_B.primary_key;
SELECT
album_id,
title,
album.artist_id,
name AS artist_name
FROM album
INNER JOIN artist on artist.artist_id = album.artist_id;
+----------+---------------------------------------+-----------+
| album_id | title | artist_id | artist_name |
|----------+---------------------------------------+-----------|
| 1 | For Those About To Rock | 1 | AC/DC |
| 4 | Let There Be Rock | 1 | AC/DC |
| 2 | Balls To The Wall | 2 | Accept |
| 3 | Restless and Wild | 2 | Accept |
+----------+---------------------------------------+-----------+
album과artist간 모든 일치 조합 반환
다중 INNER JOIN
SELECT
table_A.columnX,
table_A.columnY,
table_B.columnZ, table_C columnW
FROM table_A
INNER JOIN table_B ON table_B.foreign_key = table_A.primary_key
INNER JOIN table_C ON table_C.foreign_key = table_B.primary_key;
테이블을 조인해 봅시다!
SQL Server 입문
