laravel 如何使用架构构建器添加虚拟列？
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How to add a virtual column with the schema builder?
提问by mpen
I'm creating a table like this,
我正在创建一个这样的表，
Schema::create('booking_segments', function (Blueprint $table) {
$table->increments('id');
$table->datetime('start')->index();
$table->integer('duration')->unsigned();
$table->string('comments');
$table->integer('booking_id')->unsigned();
$table->foreign('booking_id')->references('id')->on('bookings')->onDelete('cascade');
});
But I want to add one extra column. It looks like this in raw SQL:
但我想添加一个额外的列。在原始 SQL 中看起来像这样：
ALTER TABLE booking_segments ADD COLUMN `end` DATETIME AS (DATE_ADD(`start`, INTERVAL duration MINUTE)) PERSISTENT AFTER `start`
How can I add it in my migration? I will also need to create an index on it.
如何将其添加到我的迁移中？我还需要在其上创建一个索引。
回答by Angelos Koufakis
I know this is an old question, but there is a way to do it using the schema builder since Laravel 5.3, so I thought I would put it here for completeness.
我知道这是一个老问题，但是自 Laravel 5.3 以来，有一种使用模式构建器的方法可以做到这一点，所以我想我会把它放在这里是为了完整性。
You can use laravel 5.3 column modifiersvirtualAs or storedAs.
您可以使用 laravel 5.3列修饰符virtualAs 或 storedAs。
So, to create a virtual generated column to be computed at every query you would create the column like this:
因此，要创建要在每个查询中计算的虚拟生成列，您可以像这样创建列：
$table->dateTime('created_at')->virtualAs( 'DATE_ADD(`start`, INTERVAL duration MINUTE)' );
To create a stored generated column you would create the column like this instead:
要创建存储的生成列，您可以像这样创建列：
$table->dateTime('created_at')->storedAs( 'DATE_ADD(`start`, INTERVAL duration MINUTE)' );
回答by lukasgeiter
I don't think you can do it with the schema builder (someone please correct me if I'm wrong) but you can always "fall back" to raw SQL:
我不认为你可以用模式构建器来做到这一点（如果我错了，请有人纠正我），但你总是可以“退回”到原始 SQL：
DB::statement('ALTER TABLE booking_segments ADD COLUMN `end` DATETIME AS (DATE_ADD(`start`, INTERVAL duration MINUTE)) PERSISTENT AFTER `start`');
