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MySQL：存在しない場合は制約を追加

データベースの作成スクリプトで、次のようなスクリプトを作成します。

CREATE TABLE IF NOT EXISTS `rabbits`
(
    `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
    `name` VARCHAR(255) NOT NULL,
    `main_page_id` INT UNSIGNED COMMENT 'What page is the main one',
    PRIMARY KEY (`id`),
    KEY `main_page_id` (`main_page_id`)
)
ENGINE=InnoDB;

CREATE TABLE IF NOT EXISTS `rabbit_pages`
(
    `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
    `rabbit_id` INT UNSIGNED NOT NULL,
    `title` VARCHAR(255) NOT NULL,
    `content` TEXT NOT NULL,
    PRIMARY KEY (`id`),
    KEY `rabbit_id` (`rabbit_id`),
    CONSTRAINT `fk_rabbits_pages` FOREIGN KEY (`rabbit_id`) REFERENCES `rabbits` (`id`)
)
ENGINE=InnoDB;

ALTER TABLE `rabbits`
    ADD CONSTRAINT `fk_rabbits_main_page` FOREIGN KEY (`main_page_id`) REFERENCES `rabbit_pages` (`id`);

これは最初は正常に実行されますが、再度実行すると、最後の行で「書き込みまたは更新時にキーが重複しています」と失敗します。

ADD CONSTRAINT IF NOT EXISTSなどを実行する方法はありますか？ CREATE TABLEクエリでできるように？

mysqlconstraints
19
Svish

興味深い質問です。 CREATE TABLEステートメントを呼び出す前に外部キーを無効にし、後で有効にすることをお勧めします。これにより、外部キーをCREATE TABLEDDLで直接定義できるようになります。

例：

SET FOREIGN_KEY_CHECKS = 0;
Query OK, 0 rows affected (0.00 sec)

CREATE TABLE IF NOT EXISTS `rabbits` (
    `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
    `name` VARCHAR(255) NOT NULL,
    `main_page_id` INT UNSIGNED COMMENT 'What page is the main one',
    PRIMARY KEY (`id`),
    KEY `main_page_id` (`main_page_id`),
    CONSTRAINT `fk_rabbits_main_page` FOREIGN KEY (`main_page_id`) REFERENCES `rabbit_pages` (`id`)
) ENGINE=InnoDB;
Query OK, 0 rows affected (0.04 sec)

CREATE TABLE IF NOT EXISTS `rabbit_pages` (
    `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
    `rabbit_id` INT UNSIGNED NOT NULL,
    `title` VARCHAR(255) NOT NULL,
    `content` TEXT NOT NULL,
    PRIMARY KEY (`id`),
    KEY `rabbit_id` (`rabbit_id`),
    CONSTRAINT `fk_rabbits_pages` FOREIGN KEY (`rabbit_id`) REFERENCES `rabbits` (`id`)
) ENGINE=InnoDB;
Query OK, 0 rows affected (0.16 sec)

SET FOREIGN_KEY_CHECKS = 1;
Query OK, 0 rows affected (0.00 sec)

テストケース：

INSERT INTO rabbits (name, main_page_id) VALUES ('bobby', NULL);
Query OK, 1 row affected (0.02 sec)

INSERT INTO rabbit_pages (rabbit_id, title, content) VALUES (1, 'My Main Page', 'Hello');
Query OK, 1 row affected (0.00 sec)

SELECT * FROM rabbits;
+----+-------+--------------+
| id | name  | main_page_id |
+----+-------+--------------+
|  1 | bobby | NULL         |
+----+-------+--------------+
1 row in set (0.00 sec)

SELECT * FROM rabbit_pages;
+----+-----------+--------------+---------+
| id | rabbit_id | title        | content |
+----+-----------+--------------+---------+
|  1 |         1 | My Main Page | Hello   |
+----+-----------+--------------+---------+
1 row in set (0.00 sec)

UPDATE rabbits SET main_page_id = 2 WHERE id = 1;
ERROR 1452 (23000): A foreign key constraint fails

UPDATE rabbits SET main_page_id = 1 WHERE id = 1;
Query OK, 1 row affected (0.00 sec)
Rows matched: 1  Changed: 1  Warnings: 0

UPDATE rabbit_pages SET rabbit_id = 2 WHERE id = 1;
ERROR 1452 (23000): A foreign key constraint fails
17
Daniel Vassallo

FOREIGN_KEY_CHECKSは優れたツールですが、テーブルを削除して再作成せずにこれを行う方法を知る必要がある場合。 SELECTステートメントONinformation_schema.TABLE_CONSTRAINTSを使用して、外部キーが存在するかどうかを判別できます。

IF NOT EXISTS (
    SELECT NULL 
    FROM information_schema.TABLE_CONSTRAINTS
    WHERE
        CONSTRAINT_SCHEMA = DATABASE() AND
        CONSTRAINT_NAME   = 'fk_rabbits_main_page' AND
        CONSTRAINT_TYPE   = 'FOREIGN KEY'
)
THEN
    ALTER TABLE `rabbits`
    ADD CONSTRAINT `fk_rabbits_main_page`
    FOREIGN KEY (`main_page_id`)
    REFERENCES `rabbit_pages` (`id`);
END IF
27
Paul Ostrowski

MariaDBは、この構文を 10.0.2以降 でサポートしています。

ALTER TABLE `rabbits`
ADD CONSTRAINT `fk_rabbits_main_page` FOREIGN KEY IF NOT EXISTS
(`main_page_id`) REFERENCES `rabbit_pages` (`id`);
15
seanf