views:

36

answers:

1

In my create script for my database create script looking something like this:

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`);

This runs fine the first time, but if I run it again it fails on the last line there with "Duplicate key on write or update".

Is there a way I can do sort of a ADD CONSTRAINT IF NOT EXISTS or something like that? Like I can do with the CREATE TABLE query?

+2  A: 

Interesting question. You may want to disable foreign keys before you call your CREATE TABLE statements and enable them afterwards. This will allow you to define the foreign keys directly in the CREATE TABLE DDL:

Example:

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)

Test case:

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
Daniel Vassallo
Hm, I thought I had to create table A, create table B with B->A and then add A->B because the constraint would fail if a table does not exist yet... weird... will try this out ASAP :p
Svish
@Svish: It would, unless you have that `SET FOREIGN_KEY_CHECKS = 0;` at the top. That's the trick. We're then setting it back to `1` at the end.
Daniel Vassallo
Awesome. Totally worked :D
Svish