你可能会得到外键约束错误的原因:

You are not using InnoDB as the engine on all tables. You are trying to reference a nonexistent key on the target table. Make sure it is a key on the other table (it can be a primary or unique key, or just a key) The types of the columns are not the same (an exception is the column on the referencing table can be nullable even if it is not nullable in the referenced table). If the primary key or foreign key is a varchar, make sure the collation is the same for both. One of the reasons may also be that the column you are using for ON DELETE SET NULL is not defined to be null. So make sure that the column is set default null.

检查这些。

当使用Laravel迁移时尝试创建外键时，就像下面的例子:

用户表

public function up()
{
    Schema::create('flights', function (Blueprint $table) {
        $table->increments('id');
        $table->string('name');
        $table->TinyInteger('color_id')->unsigned();
        $table->foreign('color_id')->references('id')->on('colors');
        $table->timestamps();
    });
}

颜色表

public function up()
{
    Schema::create('flights', function (Blueprint $table) {
        $table->increments('id');
        $table->string('color');
        $table->timestamps();
    });
}

有时候属性不起作用:

[PDOException]
SQLSTATE[HY000]: General error: 1215 Cannot add foreign key constraint

发生此错误是因为[用户表]中的外键(类型)与[颜色表]中的主键(类型)不同。

要解决这个问题，你应该改变[colors table]中的主键:

($ table - > tinyIncrements’id’);

当你使用主键$table->Increments('id');时，你应该使用Integer作为外键:

$table->unsignedInteger('fk_id');
$table->foreign('fk_id')->references('id')->on('table_name');

$table->tinyIncrements('id');你应该使用unsignedTinyInteger作为外键:

$table->unsignedTinyInteger('fk_id');
$table->foreign('fk_id')->references('id')->on('table_name');

当你使用主键$table->smallIncrements('id');你应该使用unsignedSmallInteger作为外键:

$table->unsignedSmallInteger('fk_id');
$table->foreign('fk_id')->references('id')->on('table_name');

当你使用主键$table->mediumIncrements('id');你应该使用unsignedMediumInteger作为外键:

$table->unsignedMediumInteger('fk_id');
$table->foreign('fk_id')->references('id')->on('table_name');

In my case, I had deleted a table using SET FOREIGN_KEY_CHECKS=0, then SET FOREIGN_KEY_CHECKS=1 after. When I went to reload the table, I got error 1215. The problem was there was another table in the database that had a foreign key to the table I had deleted and was reloading. Part of the reloading process involved changing a data type for one of the fields, which made the foreign key from the other table invalid, thus triggering error 1215. I resolved the problem by dropping and then reloading the other table with the new data type for the involved field.

对于其他人，相同的错误可能并不总是由于列类型不匹配。您可以通过发出这个命令找到关于MySQL外键错误的更多信息

SHOW ENGINE INNODB STATUS;

您可能会在打印的消息顶部附近发现一个错误。类似的

属性所在的引用表中找不到索引 引用的列显示为第一个列或列类型 在表和引用表中不匹配约束。

我猜是客户。Case_Number和/或Staff。Emp_ID与Clients_has_Staff的数据类型不完全相同。Clients_Case_Number和Clients_has_Staff.Staff_Emp_ID。

也许父表中的列是INT UNSIGNED?

两个表中的数据类型必须完全相同。

当列的类型不相同时也会发生这种情况。

例如，如果你引用的列是一个UNSIGNED INT，而被引用的列是INT，那么你会得到这个错误。