Question

在Laravel迁移中，我有以下一行：

$table->tinyInteger('action_type')->index();

并且当我尝试使用以下代码将数据保存在表中时：

ActionsLog::create([
    'account_id' => auth()->user()->id,
    'action_type' => 1,  //like
    'log', $ex->getMessage()
]);

laravel无法返回名称为action_type的第三列，并且出现此错误：

Column not found: 1054 Unknown column '1' in 'field list' (SQL: insert into `actions_logs` (`account_id`, `action_type`, `1`, `updated_at`, `created_at`) values (2, 1, SQLSTATE[42S22]: Column not found: 1054 Unknown column '1' in 'field list' (SQL: insert into `actions_logs` (`account_id`, `action_type`, `1`, `updated_at`, `created_at`) values (2, 1, Use of undefined constant ture - assumed 'ture' (this will throw an Error in a future version of PHP), 2018-07-11 07:50:18, 2018-07-11 07:50:18)), 2018-07-11 07:50:18, 2018-07-11 07:50:18))

我的迁移文件：

public function up()
{
    Schema::create('actions_logs', function (Blueprint $table) {
        $table->increments('id');
        $table->integer('account_id')->unsigned();
        $table->foreign('account_id')->references('id')->on('instagram_actions_histories')->onDelete('cascade');
        $table->tinyInteger('action_type')->index();
        $table->text('log');
        $table->timestamps();
    });
}

Answer 1

问题出在第'log', $ex->getMessage()行。将此行中的,更改为=>。将代码替换为：

ActionsLog::create([
    'account_id' => auth()->user()->id,
    'action_type' => 1,
    'log' => $ex->getMessage()
]);

Answer 2

您应该像那样创建

ActionsLog::create([
'account_id' => auth()->user()->id,
'action_type' => 1,
'log'=> $ex->getMessage()
]);

Laravel返回错误的数据库列名以保存数据

2 个答案: