我有以下两个带有外键的表status_id:
mysql> describe usr_cookbook;
+-----------+-----------+------+-----+-------------------+-------+
| Field | Type | Null | Key | Default | Extra |
+-----------+-----------+------+-----+-------------------+-------+
| user_id | int(11) | NO | PRI | NULL | |
| recipe_id | int(11) | NO | PRI | NULL | |
| status_id | int(11) | NO | MUL | NULL | |
| added_ts | timestamp | NO | | CURRENT_TIMESTAMP | |
+-----------+-----------+------+-----+-------------------+-------+
mysql> describe usr_cookbook_status;
+-----------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-----------+-------------+------+-----+---------+-------+
| status_id | int(11) | NO | PRI | NULL | |
| name | varchar(45) | NO | | NULL | |
+-----------+-------------+------+-----+---------+-------+
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中有两行usr_cookbook_status:
mysql> select * from usr_cookbook_status;
+-----------+-----------+
| status_id | name |
+-----------+-----------+
| 1 | Try Soon |
| 2 | Favorites |
+-----------+-----------+
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所以我应该很好地插入 usr_cookbook 对吧?
mysql> insert into usr_cookbook (user_id, recipe_id, status_id) values (3, 5, 1);
ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint
fails (`tomato`.`usr_cookbook`, CONSTRAINT `fk_cookbook_status` FOREIGN KEY
(`status_id`)
REFERENCES `usr_cookbook_status` (`status_id`) ON DELETE NO ACTION ON UPDATE
NO ACTION)
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为什么会违反约束?在status_id父表我使用(1)存在。我确定我在这里遗漏了一些简单的东西......
当我按照 Brian 的建议查看 DDL 时,问题立即变得清晰起来:该usr_cookbook_status表正在使用 MyISAM 引擎。我把它切换到 InnoDB,现在一切正常了。