如何将 Laravel 个迁移字段克隆到另一个迁移中?

How to clone Laravel migration fields into another migration?

我有这个 Laravel 迁移结构:

class CreateWarehouseProductTable extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {
        Schema::create('warehouse_products', function (Blueprint $table) {
            $table->bigIncrements('id');
            $table->integer('product_id')->default(0);
            $table->integer('warehouse_id');
            $table->integer('free_amount')->default(0);
            $table->integer('booked_amount')->default(0);
            // ...
            $table->timestamps();
        });
    }

    // ...
}

我需要从仓库产品创建一个每日备份,为此我需要创建一个新的 table,它与 warehouse_products 完全相同,并且包含一个额外的备份日期字段。

有什么最佳做法吗?我想是这样的:

class CreateWarehouseProductBackupTable extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {
        Schema::create('warehouse_product_backups', function (Blueprint $table) {
            CreateWarehouseProductTable->cloneFieldsToHere();
            $table->date('date_of_backup');
        });
    }
}

从现有迁移中克隆字段是否有类似的好的做法?

我找到了一个解决方案,我认为它并不优雅,但它有效:

class CreateWarehouseProductTable extends Migration
{
    public function up()
    {
        Schema::create('warehouse_products', function (Blueprint $table) {
            $this->setWarehouseProductColumns($table);
        });

        Schema::create('warehouse_product_backups', function (Blueprint $table) {
            $this->setWarehouseProductColumns($table);
            $table->date('date_of_backup')->nullable();
        });
    }

    public function setWarehouseProductColumns(Blueprint $table) {
            $table->bigIncrements('id');
            $table->integer('product_id')->default(0);
            $table->integer('warehouse_id');
            $table->integer('free_amount')->default(0);
            $table->integer('booked_amount')->default(0);
            // ...
            $table->timestamps();
    }
}