Question

刚刚在mac os x 10.6上安装了MySQL 5.5，并且在许多表上遇到了一个奇怪的问题。以下是一个例子。插入行失败时带有外键约束，但不应该。它引用的外键确实存在。有什么想法吗？

mysql> show create table Language;
+----------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| Table    | Create Table                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               |
+----------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| Language | CREATE TABLE `Language` (
  `Id` int(11) NOT NULL AUTO_INCREMENT,
  `Code` varchar(2) NOT NULL,
  `Name` varchar(63) CHARACTER SET utf8 DEFAULT NULL,
  `Variant` varchar(63) CHARACTER SET utf8 DEFAULT NULL,
  `Country_Id` int(11) DEFAULT NULL,
  PRIMARY KEY (`Id`),
  UNIQUE KEY `Code` (`Code`,`Country_Id`,`Variant`),
  KEY `FKA3ACF7789C1796EB` (`Country_Id`),
  CONSTRAINT `FKA3ACF7789C1796EB` FOREIGN KEY (`Country_Id`) REFERENCES `Country` (`Id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1 |
+----------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)

mysql> show create table Language_Phrases;
+------------------+-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| Table            | Create Table                                                                                                                                                                                                                                                                                                                                                    |
+------------------+-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| Language_Phrases | CREATE TABLE `Language_Phrases` (
  `Language_Id` int(11) NOT NULL,
  `Phrase` varchar(255) DEFAULT NULL,
  `Label` varchar(255) NOT NULL,
  PRIMARY KEY (`Language_Id`,`Label`),
  KEY `FK8B4876F3AEC1DBE9` (`Language_Id`),
  CONSTRAINT `FK8B4876F3AEC1DBE9` FOREIGN KEY (`Language_Id`) REFERENCES `Language` (`Id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 |
+------------------+-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)

mysql> select * from Language;
+----+------+----------+---------+------------+
| Id | Code | Name     | Variant | Country_Id |
+----+------+----------+---------+------------+
|  1 | en   | English  |         |        235 |
|  2 | ro   | Romanian |         |        181 |
+----+------+----------+---------+------------+
2 rows in set (0.00 sec)

mysql> select * from Language_Phrases;
Empty set (0.00 sec)

mysql> INSERT INTO Language_Phrases (Language_Id, Label, Phrase) VALUES (1, 'exampleLabel', 'Some phrase');
ERROR 1452 (23000): Cannot add or update a child row: a foreign key constraint fails (`dev`.`language_phrases`, CONSTRAINT `FK8B4876F3AEC1DBE9` FOREIGN KEY (`Language_Id`) REFERENCES `Language` (`Id`))
mysql>

更新：在多次删除并重新创建数据库之后，我在上面的插入失败后执行了show engine innodb status并获得了令人惊讶的结果。找不到父语言表！这看起来很奇怪......任何想法？

------------------------
LATEST FOREIGN KEY ERROR
------------------------
110406  9:55:49 Transaction:
TRANSACTION CA3B, ACTIVE 0 sec, OS thread id 4494462976 inserting
mysql tables in use 1, locked 1
1 lock struct(s), heap size 376, 0 row lock(s)
MySQL thread id 25, query id 50720 localhost root update
INSERT INTO Language_Phrases (Language_Id, Label, Phrase) VALUES (1, 'exampleLabel', 'Some phrase')
Foreign key constraint fails for table `dev`.`language_phrases`:
,
  CONSTRAINT `FK8B4876F3AEC1DBE9` FOREIGN KEY (`Language_Id`) REFERENCES `Language` (`Id`)
Trying to add to index `PRIMARY` tuple:
DATA TUPLE: 5 fields;
 0: len 4; hex 80000001; asc     ;;
 1: len 17; hex 747970654d69736d617463682e79656172; asc exampleLabel;;
 2: len 6; hex 00000000ca3b; asc      ;;;
 3: len 7; hex 00000000000000; asc        ;;
 4: len 21; hex 59656172206d7573742062652061206e756d626572; asc Some phrase;;

But the parent table `dev`.`Language`
or its .ibd file does not currently exist!

更新2 ：事实证明这只是MySQL中的大量错误。显然MySQL的最新版本在mac os X 10.6下可能无法完全运行（可能也是早期的版本？）。降级至5.5.8似乎有效。非常令人惊讶。

Answer 1

这似乎是自Mac OS X上的MySQL 5.5.9以来引入的错误： http://bugs.mysql.com/bug.php?id=60309

在5.5.13（5月31日发布）中标记为已修复并在发行说明中提及： http://dev.mysql.com/doc/refman/5.5/en/news-5-5-13.html

或者，我在5.5.10上验证过的错误报告中列出了一种解决方法，并在下面转载：


[20 Mar 11:29] Harald Neiss

I also received a new MBP and reinstalled MySQL (mysql-5.5.10-osx10.6-x86_64). Finally I
came across the same problem as described above. So here is the query result and what I
did to solve it.

mysql> show variables like 'lower%';
+------------------------+-------+
| Variable_name          | Value |
+------------------------+-------+
| lower_case_file_system | ON    |
| lower_case_table_names | 2     |
+------------------------+-------+
2 rows in set (0.00 sec)

Dropped database, created the file /etc/my.cnf with the following content:

[mysqld]
lower_case_table_names=1

Restarted the MySQL daemon and repeated the query:

mysql> show variables like 'lower%';
+------------------------+-------+
| Variable_name          | Value |
+------------------------+-------+
| lower_case_file_system | ON    |
| lower_case_table_names | 1     |
+------------------------+-------+
2 rows in set (0.00 sec)

I recreated the tables and everything works fine.

Answer 2

恕我直言，这并不奇怪。我在MySQL中发现了许多错误。例如，使用where子句运行查询，例如＆＃34; WHERE some_tinyint_column = 0＆＃34;它应该不会产生任何数据，但重写该条款为＆＃34; WHERE（不是some_tinyint_column = 1）＆＃34;产生结果。经过一些研究后，我发现这是一个应该修复过的bug，但是在我使用的版本中，bug仍然存在。

结论：当某些东西在MySQL中完全没有意义时，我通常会发现它可以安全地假设它是一个错误，并开始研究这些信息。

Answer 3

检查Language_Phrases (Language_Id)和Language（Id）

的数字类型属性

两者都应该是 UNSIGNED ZEROFILL 或 SIGNED

Answer 4

* MySQL的＆GT; INSERT INTO Language_Phrases（Language_Id，Label，Phrase）VALUES（1，'exampleLabel'，'some phrase'）; 错误1452（23000）：无法添加或更新子行：... *

您正在尝试将 1 作为Language_Id插入，但表格语言具有属性AUTO_INCREMENT = 3 。在这种情况下，您应该使用3或更高。

Answer 5

我今天遇到了同样的错误。在我的例子中，我使用了一个脚本来重新创建一些包含所有记录的表。

事实上，我已经意识到我的表之间的“引擎”类型不同：一个是MyISAM，第二个（FK的引用）是InnoDB。我把所有的表都改成了InnoDB，现在一切正常。

此脚本将生成更新脚本文件（Reference）

mysql -u DB_USER -pDB_PASSWORD --default-character-set=utf8  DATABASE_NAME -e "SELECT CONCAT('ALTER TABLE ', table_name, ' ENGINE=InnoDB;') AS sql_statements FROM information_schema.tables AS tb WHERE table_schema = database() AND ENGINE = 'MyISAM' AND TABLE_TYPE = 'BASE TABLE' ORDER BY table_name DESC;" > ./alter_InnoDb.sql

您必须删除“alter_InnoDb.sql”中的第一行，该行包含文本“sql_statements”。

之后，您可以在数据库中执行脚本来更正此错误：

mysql -u DB_USER -pDB_PASSWORD --default-character-set=utf8  DATABASE_NAME < ./ alter_InnoDb.sql

Answer 6

我也面临着同样的问题，没有找到解决我问题的解决方案。因此，由于多种原因可能会发生此问题。我只是想将所有这些原因和解决方案放在一个地方，并且还要提供有助于解决此问题的解决方案。我希望这会对以后的人有所帮助。

1）由Penfold提供-表名区分大小写
2）父子表中的引擎不匹配
3）父子表中的字符集不匹配
4）Parent（Id）和Child（Patent_Id：两者都必须具有完全相同的数据类型（也已签名/未签名）
5）InnoDB丢失了表，但是表存在-Please find solution here

SQL查询（显示引擎InnoDB状态）给出此错误（其.ibd文件当前不存在！）引用

6）很少有较早版本的mysql有此错误。Bug#60196，Bug#60309
7）这对我有用（如果上面的一切似乎还不错，您可能需要这样做）-从子表中删除外键并再次添加约束。如果它也失败，则删除父表并再次创建它，在此之前，您需要从所有子表中删除外键。这是最后一件事。
进一步阅读：
https://dev.mysql.com/doc/refman/8.0/en/server-system-variables.html#sysvar_lower_case_table_names

当外键存在时，MySQL 5.5外键约束失败

6 个答案: