更改表时出错,添加约束外键获取错误“无法添加或更新子行"
mysql> DESCRIBE questions;
+----------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------+--------------+------+-----+---------+----------------+
| id | int(255) | NO | PRI | NULL | auto_increment |
| question | varchar(255) | NO | | NULL | |
| type | char(1) | YES | | NULL | |
+----------+--------------+------+-----+---------+----------------+
mysql> DESCRIBE answers;
+--------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+--------------+------+-----+---------+----------------+
| id | int(255) | NO | PRI | NULL | auto_increment |
| answer | varchar(255) | NO | | NULL | |
| questionid | int(255) | NO | | NULL | |
| questions_id | int(255) | NO | | NULL | |
+--------------+--------------+------+-----+---------+----------------+
我正在使用这种说法:
ALTER TABLE 答案添加外键(questions_id)参考问题(id);
ALTER TABLE answers ADD FOREIGN KEY(questions_id) REFERENCES questions(id);
但我得到这个错误:
ERROR 1452 (23000):无法添加或更新子行:外键约束失败(surveydb
.#sql-df_32
,CONSTRAINT #sql-df_32_ibfk_1
FOREIGN KEY (questions_id
) REFERENCES questions
(id
)) 到您的 MySQL 服务器版本,以便在附近使用正确的语法第 1 行的描述问题"
ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint fails (
surveydb
.#sql-df_32
, CONSTRAINT#sql-df_32_ibfk_1
FOREIGN KEY (questions_id
) REFERENCESquestions
(id
))to your MySQL server version for the right syntax to use near 'DESCREBE questions' at line 1
推荐答案
answers.questions_id
中至少有一个数据值在 questions.id
中没有出现.
You have at least one data value in answers.questions_id
that does not occur in questions.id
.
这是我的意思的一个例子:
Here's an example of what I mean:
mysql> create table a ( id int primary key);
mysql> create table b ( aid int );
mysql> insert into a values (123);
mysql> insert into b values (123), (456);
mysql> alter table b add foreign key (aid) references a(id);
ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint
fails (`test`.`#sql-3dab_e5c`, CONSTRAINT `#sql-3dab_e5c_ibfk_1` FOREIGN KEY
(`aid`) REFERENCES `a` (`id`))
您可以使用它来确认存在不匹配的值:
You can use this to confirm that there are unmatched values:
SELECT COUNT(*)
FROM answers AS a
LEFT OUTER JOIN questions AS q ON a.questions_id = q.id
WHERE q.id IS NULL
相关文章