[英]SQLSTATE[HY000]: General error: 1005 Can't create table `school`.`posts` (errno: 150 "Foreign key constraint is incorrectly formed")
[英]SQLSTATE[HY000]: General error: 1005 Can't create table `test`.`members` (errno: 150 “Foreign key constraint is incorrectly formed”)
我在 Laravel 中使用我的迁移来创建表之间的关系,我有 4 个表:用户、成员、成员技能和技能。 我有用户表的以下代码:
public function up()
{
Schema::create('users', function (Blueprint $table) {
$table->id();
$table->string('name');
$table->string('email')->unique();
$table->timestamp('email_verified_at')->nullable();
$table->string('password');
$table->rememberToken();
$table->timestamps();
$table->boolean('admin');
});
}
成员表:
public function up()
{
Schema::create('members', function (Blueprint $table) {
$table->id();
$table->timestamps();
$table->string('name');
$table->string('status');
$table->date('date')->nullable();
$table->text('project')->nullable();
$table->date('start')->nullable();
$table->foreign('name')->references('name')->on('users');
});
}
member_skills 表:
public function up()
{
Schema::create('member_skills', function (Blueprint $table) {
$table->id();
$table->timestamps();
$table->string('name');
$table->string('skill');
$table->foreign('name')->references('name')->on('members');
});
}
和技能表:
public function up()
{
Schema::create('skills', function (Blueprint $table) {
$table->id();
$table->timestamps();
$table->string('skill');
$table->text('description');
$table->foreign('skill')->references('skill')->on('member_skills');
});
}
但是,运行我的迁移结果为(errno: 150 "Foreign key constraint is incorrectly formed")
。 我已经读过更改迁移顺序应该可以解决问题,所以我按照用户、成员、成员技能和技能的顺序排列了要迁移的 4 个表,但我仍然收到同样的错误。 还有什么我做错了吗?
这是执行此操作的正确方法
public function up()
{
Schema::create('members', function (Blueprint $table) {
...
$table->unsignedBigInteger('user_id');
$table->foreign('user_id')->references('id')->on('users');
});
}
public function up()
{
Schema::create('member_skills', function (Blueprint $table) {
...
$table->unsignedBigInteger('member_id');
$table->foreign('member_id')->references('id')->on('members');
});
}
public function up()
{
Schema::create('skills', function (Blueprint $table) {
...
$table->unsignedBigInteger('member_skill_id');
$table->foreign('member_skill_id')->references('id')->on('member_skills');
});
}
更多:https://laravel.com/docs/8.x/migrations#foreign-key-constraints
您应该尝试使用成员表的id作为外键,而不是 member_skills 模式中的名称
public function up()
{
Schema::create('member_skills', function (Blueprint $table) {
$table->id();
$table->timestamps();
$table->string('member_id');
$table->string('skill');
$table->foreign('member_id')->references('id')->on('members');
});
}
您收到此错误是因为您试图在成员表中引用名称,该成员表已经是用户表的外键。
您可以通过刀片中的id外键访问成员的名称。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.