错误更改表，添加约束外键获取错误“无法添加或更新子行” [英] error altering table, adding constraint foreign key getting error "Cannot add or update a child row"

问题描述

 的MySQL>描述问题; 
 + ---------- + -------------- + ------ + ----- + ------ --- + ---------------- + 
 |字段|类型|空|密钥|默认|额外| 
 + ---------- + -------------- + ------ + ----- + ------ --- + ---------------- + 
 | id | int（255）| NO | PRI | NULL | auto_increment | 
 |问题| varchar（255）| NO | | NULL | | 
 |键入| char（1）|是| | NULL | | 
 + ---------- + -------------- + ------ + ----- + ------ --- + ---------------- + 
 mysql>描述答案; 
 + -------------- + -------------- + ------ + ----- +  - ------- + ---------------- + 
 |字段|类型|空|密钥|默认|额外| 
 + -------------- + -------------- + ------ + ----- +  - ------- + ---------------- + 
 | id | int（255）| NO | PRI | NULL | auto_increment | 
 |回答| varchar（255）| NO | | NULL | | 
 | questionid | int（255）| NO | | NULL | | 
 | questions_id | int（255）| NO | | NULL | | 
 + -------------- + -------------- + ------ + ----- +  - ------- + ---------------- + 
  

我使用这个语句：

ALTER TABLE答案ADD FOREIGN KEY（questions_id）参考题目（id）;

但我得到这个错误：

错误1452（23000）：Can not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 ）参考疑问（<$

<您在 answers.questions_id 中至少有一个数据值不会出现在 questions.id 。

下面是我的意思的例子：

 的MySQL>创建表a（id int主键）; 
 
 mysql>创建表b（aid int）; 
 
 mysql>插入一个值（123）; 
 
 mysql>插入b值（123），（456）; 
 
 mysql> alter table b add foreign key（aid）引用a（id）; 
错误1452（23000）：无法添加或更新子行：外键约束
失败（`test`.`＃sql-3dab_e5c`，CONSTRAINT`＃sql-3dab_e5c_ibfk_1` FOREIGN KEY 
（`aid`）参考`a`（`id`））
  

你可以使用这个确认有不匹配的值：

pre $ $ $ $ $ $ $ OUTER JOIN问题AS q ON a.questions_id = q.id
WHERE q.id IS NULL

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

I am using this statement:

ALTER TABLE answers ADD FOREIGN KEY(questions_id) REFERENCES questions(id); 

but i get this error:

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) REFERENCES questions (id))to your MySQL server version for the right syntax to use near 'DESCREBE questions' at line 1

解决方案

You have at least one data value in answers.questions_id that does not occur in questions.id.

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`))

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

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