违反完整性约束。没有矛盾。 Laravel

Question

我该如何解决这个错误？我似乎找不到违反完整性约束的情况。

错误：

Illuminate\Database\QueryException: SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot add or update a child row: a foreign key constraint fails (kondate.created_kondates_foods, CONSTRAINT created_kondates_foods_food_id_foreign FOREIGN KEY (food_id) REFERENCES foods (id)) (SQL: insert into created_kondates_foods (created_kondate_id, food_id, updated_at, created_at) values (34, 6655, 2022-04-05 19:00:31, 2022-04-05 19:00:31)) in file /Users/Developments/kondate/vendor/laravel/framework/src/Illuminate/Database/Connection.php on line 669

我的代码：

class CreatedKondate extends Model
{
    protected $connection = 'kondate';

    protected $table = 'created_kondates';

    protected $guarded = ['id'];

    public function foods()
    {
        return $this->belongsToMany('App\Models\Food', 'created_kondates_foods', 'created_kondate_id', 'food_id');
    }
}

中级模型table‖.

CreatedKondateFood

class CreatedKondateFood extends Model
{
    protected $connection = 'kondate';

    protected $table = 'created_kondates_foods';

    protected $guarded = ['id'];

}

数据库输出函数

 $first_date = date("2022-04-01");
 $end_date = date("2022-04-t");

        for($i = $first_date; $i<=$end_date ;$i++)
        {
            $createdKondate = new CreatedKondate;
            $createdKondate->company_id = $company_id;
            $createdKondate->calendar_date = $i;
            $createdKondate->save();

        $createdKondateFood = new CreatedKondateFood;
        $createdKondateFood->created_kondate_id = $createdKondate->id;
        $createdKondateFood->food_id = '6655';
        $createdKondateFood->save();
        }
    }

Answer 1

如果换掉会怎样

    $createdKondateFood->food_id = '6655';

来自

    $createdKondateFood->food_id = 6655;

也可能是因为食物中没有id设置为6655的物品table

违反完整性约束。没有矛盾。 Laravel

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