php larvel 加入所有表时删除多个表中的数据

Question

我的代码有问题。我必须使用 id 从 5 个表中删除数据，这些表由主键和外键连接。

这是我尝试过的，但它会显示添加订阅表 ID。但是已经添加了订阅表 id。

$subscription = DB::table('tbl_asset_subscription')->where('id',$post['asset_id'])->get();

        foreach($subscription as $row)

        {

        DB::table('tbl_asset_subscription')->where('id',$row->id)->delete();

        }

        $orderId = array();

        foreach($subscription as $row)

        {

        $order = DB::table('tbl_asset_order')->where('subscription_id',$row->id)->first();

        $orderId[] = $order->id;

        }

        foreach($orderId as $row)

        {

        DB::table('tbl_asset_payment')->where('order_id',$row->id)->delete();

        DB::table('tbl_asset_order')->where('id',$row->id)->delete();

        }

        DB::table('tbl_asset_versions')->where('asset_id',$post['asset_id'])->delete();

        DB::table('tbl_assets')->where('id',$post['asset_id'])->delete();

        // DB::table('tbl_asset_subscription')->where('asset_id',$post['asset_id'])->delete();

        echo(json_encode(array("result" => true)));

{

    "message": "SQLSTATE[23000]: Integrity constraint violation: 1451 Cannot delete or update a parent row: a foreign key constraint fails (`piccoscript`.`tbl_asset_subscription`, CONSTRAINT `fk_tbl_asset_subscription_tbl_assets` FOREIGN KEY (`asset_id`) REFERENCES `tbl_assets` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION) (SQL: delete from `tbl_assets` where `id` = 1)",

    "exception": "Illuminate\\Database\\QueryException",

    "file": "/var/www/html/piccoscript/vendor/laravel/framework/src/Illuminate/Database/Connection.php",

    "line": 664,

    "trace": [

        {

            "file": "/var/www/html/piccoscript/vendor/laravel/framework/src/Illuminate/Database/Connection.php",

            "line": 624,

            "function": "runQueryCallback",

            "class": "Illuminate\\Database\\Connection",

            "type": "->"

        },

Answer 1

您可以暂时禁用外键约束在 Laravel 中，您可以执行以下操作：$orderId = array();$subscription = DB::table('tbl_asset_subscription')->where('id',$post['asset_id'])->get(); DB::transaction(function() {    DB::statement('SET FOREIGN_KEY_CHECKS=0');    foreach($subscription as $row) {        DB::table('tbl_asset_subscription')->where('id', $row->id)->delete();    }    foreach($subscription as $row) {        $order = DB::table('tbl_asset_order')->where('subscription_id', $row->id)->first();        $orderId[] = $order->id;    }    if($orderId) {        DB::table('tbl_asset_payment')->whereIn('order_id', $orderId)->delete();        DB::table('tbl_asset_order')->whereIn('id', $orderId )->delete();    }    DB::table('tbl_asset_versions')->where('asset_id', $post['asset_id'])->delete();    DB::table('tbl_assets')->where('id', $post['asset_id'])->delete();    DB::statement('SET FOREIGN_KEY_CHECKS=1');});echo(json_encode(array("result" => true)));

Answer 2

$orderId 不是关联数组，因此出现错误。试试这个代码。foreach($subscription as $row){$order = DB::table('tbl_asset_order')->where('subscription_id',$row->id)->first();$orderId[] = array(        'id' => $order->id    );}foreach($orderId as $row){DB::table('tbl_asset_payment')->where('order_id',$row->id)->delete();DB::table('tbl_asset_order')->where('id',$row->id)->delete();}DB::table('tbl_asset_versions')->where('asset_id',$post['asset_id'])->delete();DB::table('tbl_assets')->where('id',$post['asset_id'])->delete();// DB::table('tbl_asset_subscription')->where('asset_id',$post['asset_id'])->delete();echo(json_encode(array("result" => true)));