mysql - MySQL 中的外键添加失败，错误代码为 1005，编号为 150

Question

所以我正在尝试向我的一个表中添加一个新的外键:

 ALTER TABLE `UserTransactions`.`ExpenseBackTransactions` 
   ADD CONSTRAINT `FK_EBTx_CustomAccountID`
   FOREIGN KEY (`CustomAccountID` )
   REFERENCES `UserTransactions`.`CustomAccounts` (`CustomAccountID`)
   ON DELETE NO ACTION
   ON UPDATE NO ACTION,
   ADD INDEX `FK_EBTx_CustomAccountID` (`CustomAccountID` ASC) ;

我不断收到以下错误:

Error Code: 1005
Can't create table './UserTransactions/#sql-187a_29.frm' (errno: 150)

过去我对这个表和其他表进行了相当多的更改，这是我第一次遇到这个问题。有什么想法是什么原因造成的吗？

更新

我的 SHOW INNODB STATUS 错误:

------------------------
LATEST FOREIGN KEY ERROR
------------------------
110525 15:56:36 Error in foreign key constraint of table UserTransactions/#sql-187a_2c:

  FOREIGN KEY (`CustomAccountID` )
  REFERENCES `UserTransactions`.`CustomAccounts` (`CustomAccountID` )
  ON DELETE NO ACTION
  ON UPDATE NO ACTION
, ADD INDEX `FK_EBTx_CustomAccountID` (`CustomAccountID` ASC):
Cannot resolve table name close to:
 (`CustomAccountID` )
  ON DELETE NO ACTION
  ON UPDATE NO ACTION
, ADD INDEX `FK_EBTx_CustomAccountID` (`CustomAccountID` ASC)

Answer 1

有一个 nice checklist here .

Below is a running list of known causes that people have reported for the dreaded errno 150:

1. The two key fields type and/or size is not an exact match. For example, if one is INT(10) the key field needs to be INT(10) as well and not INT(11) or TINYINT. You may want to confirm the field size using SHOW CREATE TABLE because Query Browser will sometimes visually show just INTEGER for both INT(10) and INT(11). You should also check that one is not SIGNED and the other is UNSIGNED. They both need to be exactly the same. (More about signed vs unsigned here).
2. One of the key field that you are trying to reference does not have an index and/or is not a primary key. If one of the fields in the relationship is not a primary key, you must create an index for that field. (thanks to Venkatesh and Erichero and Terminally Incoherent for this tip)
3. The foreign key name is a duplicate of an already existing key. Check that the name of your foreign key is unique within your database. Just add a few random characters to the end of your key name to test for this. (Thanks to Niels for this tip)
4. One or both of your tables is a MyISAM table. In order to use foreign keys, the tables must both be InnoDB. (Actually, if both tables are MyISAM then you won’t get an error message – it just won’t create the key.) In Query Browser, you can specify the table type.
5. You have specified a cascade ON DELETE SET NULL, but the relevant key field is set to NOT NULL. You can fix this by either changing your cascade or setting the field to allow NULL values. (Thanks to Sammy and J Jammin)
6. Make sure that the Charset and Collate options are the same both at the table level as well as individual field level for the key columns. (Thanks to FRR for this tip)
7. You have a default value (ie default=0) on your foreign key column (Thanks to Omar for the tip)
8. One of the fields in the relationship is part of a combination (composite) key and does not have its own individual index. Even though the field has an index as part of the composite key, you must create a separate index for only that key field in order to use it in a constraint. (Thanks to Alex for this tip)
9. You have a syntax error in your ALTER statement or you have mistyped one of the field names in the relationship (Thanks to Christian & Mateo for the tip)
10. The name of your foreign key exceeds the max length of 64 chars. (Thanks to Nyleta for the tip)