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`);

第一次运行正常，但是如果我再次运行它会在最后一行失败并显示“写入或更新时重复键”。

有没有办法可以做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

FOREIGN_KEY_CHECKS是一个很棒的工具，但如果您需要知道如何在不丢弃和重新创建表的情况下执行此操作。您可以在SELECT上使用information_schema.TABLE_CONSTRAINTS语句来确定是否存在外键：

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 2

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

示例：

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)

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

Answer 3

MariaDB在10.0.2 or later中支持此语法：

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

MySQL：如果不存在则添加约束

3 个答案: