我有两个表，table1是一个列ID的父表，table2是一个列IDFromTable1(不是实际的名称)，当我把一个FK IDFromTable1到ID在table1，我得到的错误外键约束是不正确形成的错误。我想删除表2记录，如果表1记录被删除。谢谢你的帮助
ALTER TABLE `table2`
ADD CONSTRAINT `FK1`
FOREIGN KEY (`IDFromTable1`) REFERENCES `table1` (`ID`)
ON UPDATE CASCADE
ON DELETE CASCADE;
如果还需要其他信息，请告诉我。我是mysql的新手
当前回答
甚至我在mysql和libase上也遇到了同样的问题。 这就是问题所在: 要引用其他表的列的表在数据类型或数据类型大小方面都不同。
Error appears in below scenario:
Scenario 1:
Table A has column id, type=bigint
Table B column referenced_id type varchar(this column gets the value from the id column of Table A.)
Liquibase changeset for table B:
<changeset id="XXXXXXXXXXX-1" author="xyz">
<column name="referenced_id" **type="varchar"**>
</column>
</changeset>
<changeSet id="XXXXXXXXXXX-2" author="xyz">
<addForeignKeyConstraint constraintName="FK_table_A"
referencedTableName="A" **baseColumnNames="referenced_id**"
referencedColumnNames="id" baseTableName="B" />
</changeSet>
Table A changeSet:
<changeSet id="YYYYYYYYYY" author="xyz">
<column **name="id"** **type="bigint"** autoIncrement="${autoIncrement}">
<constraints primaryKey="true" nullable="false"/>
</column>
</changeSet>
Solution:
correct the type of table B to bigint because the referenced table has type bigint.
Scenrario 2:
The type might be correct but the size might not.
e.g. :
Table B : referenced column type="varchar 50"
Table A : base column type ="varchar 255"
Solution change the size of referenced column to that of base table's column size.
其他回答
我刚才也遇到了同样的问题。在我的例子中，我所要做的就是确保我在外键中引用的表必须在当前表之前创建(在代码的前面)。因此，如果你引用一个变量(x*5)，系统应该知道x是什么(x必须在前面的代码行中声明)。这解决了我的问题，希望它能帮助其他人。
我也有同样的问题，但我解决了。
只要确保列'ID'在'table1'有唯一的索引!
当然，这两个表中'ID'和'IDFromTable1'的列的类型和长度必须相同。但是你已经知道了。
甚至我在mysql和libase上也遇到了同样的问题。 这就是问题所在: 要引用其他表的列的表在数据类型或数据类型大小方面都不同。
Error appears in below scenario:
Scenario 1:
Table A has column id, type=bigint
Table B column referenced_id type varchar(this column gets the value from the id column of Table A.)
Liquibase changeset for table B:
<changeset id="XXXXXXXXXXX-1" author="xyz">
<column name="referenced_id" **type="varchar"**>
</column>
</changeset>
<changeSet id="XXXXXXXXXXX-2" author="xyz">
<addForeignKeyConstraint constraintName="FK_table_A"
referencedTableName="A" **baseColumnNames="referenced_id**"
referencedColumnNames="id" baseTableName="B" />
</changeSet>
Table A changeSet:
<changeSet id="YYYYYYYYYY" author="xyz">
<column **name="id"** **type="bigint"** autoIncrement="${autoIncrement}">
<constraints primaryKey="true" nullable="false"/>
</column>
</changeSet>
Solution:
correct the type of table B to bigint because the referenced table has type bigint.
Scenrario 2:
The type might be correct but the size might not.
e.g. :
Table B : referenced column type="varchar 50"
Table A : base column type ="varchar 255"
Solution change the size of referenced column to that of base table's column size.
试着运行以下命令:
show create table Parent //and check if type for both tables are the same, like myISAM or innoDB, etc //Other aspects to check with this error message: the columns used as foreign keys must be indexed, they must be of the same type (if i.e one is of type smallint(5) and the other of type smallint(6), it won't work), and, if they are integers, they should be unsigned. //or check for charsets show variables like "character_set_database"; show variables like "collation_database"; //edited: try something like this ALTER TABLE table2 ADD CONSTRAINT fk_IdTable2 FOREIGN KEY (Table1_Id) REFERENCES Table1(Table1_Id) ON UPDATE CASCADE ON DELETE CASCADE;
我在使用HeidiSQL时也遇到了同样的问题。您收到的错误非常神秘。我的问题是外键列和引用列的类型或长度不相同。
外键列是SMALLINT(5) UNSIGNED，引用列是INT(10) UNSIGNED。一旦我将它们都设置为完全相同的类型，外键的创建就会完美地工作。
