外键违规 - 不知道为什么

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    |       |
+-----------+-------------+------+-----+---------+-------+

中有两行usr_cookbook_status：

mysql> select * from usr_cookbook_status;
+-----------+-----------+
| status_id | name      |
+-----------+-----------+
|         1 | Try Soon  |
|         2 | Favorites |
+-----------+-----------+

所以我应该很好地插入 usr_cookbook 对吗？

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)

为什么违反约束？status_id我正在使用的(1) 存在于父表中。我敢肯定这里缺少一些简单的东西...