MySQL：如果不存在则添加约束

Question

在我的数据库创建脚本中，创建脚本如下所示：

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`);

第一次运行良好，但如果我再次运行它，它会在最后一行出现“写入或更新时复制键”失败。

有没有办法我可以做ADD CONSTRAINT IF NOT EXISTS 或类似的事情？就像我可以使用 CREATE TABLE 查询一样？

Answer 1

对于非 MariaDB，这对我有用：

SET @dbname = DATABASE();
SET @tablename = "my_table";
SET @constraintname = "my_constraint_name";
SET @columnname = "my_column";
SET @othertablename = "other_table";
SET @othercolumnname = "other_column_name";
SET @deleteaction = "CASCADE";
SET @updateaction = "RESTRICT";
SET @preparedStatement = (SELECT IF(
  (
    SELECT COUNT(*) FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS
    WHERE
      (table_name = @tablename)
      AND (table_schema = @dbname)
      AND (constraint_name = @constraintname)
  ) > 0,
  "SELECT 1",
  CONCAT("ALTER TABLE ",@tablename,
    " ADD CONSTRAINT ",@constraintname,
    " FOREIGN KEY(",@columnname,")
      REFERENCES ",@othertablename,"(",@othercolumnname,")
      ON DELETE ",@deleteaction,
    " ON UPDATE ",@updateaction)));
PREPARE alterIfNotExists FROM @preparedStatement;
EXECUTE alterIfNotExists;
DEALLOCATE PREPARE alterIfNotExists;

这是一个解决方案，由类似的问题制成：https://stackoverflow.com/a/31989541/3589448

根据需要添加参数。 @deleteaction 和 @updateaction 可以有：“RESTRICT”、“CASCADE”、“SET NULL”或“NO ACTION”。

Answer 2

FOREIGN_KEY_CHECKS 是一个很棒的工具，但如果您需要知道如何在不删除和重新创建表的情况下执行此操作。您可以使用SELECT 语句 ON information_schema.TABLE_CONSTRAINTS 来确定外键是否存在：

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

Answer 3

MariaDB 在10.0.2 or later 中支持这种语法：

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

Answer 4

有趣的问题。您可能希望在调用 CREATE TABLE 语句之前禁用外键，然后再启用它们。这将允许您直接在 CREATE TABLE DDL 中定义外键：

例子：

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)

测试用例：

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