如何在Laravel中获取插入行的ID？

Question

这是我的代码和错误：

正如您所看到的，我已经测试了if ( $inserted )之类的插入语句，但id没有进入该对象。我该如何解决？

这是我的代码：

    public function register_guarantee_ticket()
    {
        $problem = json_decode($_COOKIE['guarantee_ticket']);

        $guarantee_tickets = new guarantee_tickets;
        $guarantee_tickets->unique_product_id = $problem->unique_product_id;
        $guarantee_tickets->user_id = Auth::user()->id;
        $guarantee_tickets->submit_time = time();
        $guarantee_tickets->brick = !empty($problem->turn_on) ? 1 : 0;
        $guarantee_tickets->title = $problem->title;
        $guarantee_tickets->description = $problem->description;
        $guarantee_tickets->tracking_code = DB::select("SELECT LEFT(UUID(), 8) as random_unique_string;")[0]->random_unique_string;
        $guarantee_tickets->status_id = 1; // waiting ...
        $inserted = $guarantee_tickets->save();

        if ($inserted){
//            echo "ticket registered";
            $ticket_id = $guarantee_tickets->id;
            return \redirect()->route('list_of_ticket')->with($ticket_id);
            unset($_COOKIE['guarantee_ticket']);
            setcookie('guarantee_ticket', null, -1, '/');
            // redirect to user's panel, in the ticket list
        } else {
            $status = "register-unique-device-alert-danger";
            $msg = "something went wrong, try again";
            return \redirect()->route('form_serial_number')->with($status, $msg);
        }
    }

Answer 1

$guarantee_tickets->save();
$inserted = $guarantee_tickets->id;

这样就可以了。

Answer 2

你正在尝试一种不好的方法来使这个东西工作你可以直接访问id作为模型对象的变量来保存价值，这是＆＃39; $ guarantee_tickets＆＃39;

并访问id：

$id = $guarantee_tickets->id;

因为$ inserted变量

中没有任何内容

Answer 3

首先，您需要检查inserted是什么，因为您在if语句中检查它，但它可能是一个不会返回true或false的对象。

Answer 4

在public $fillable = [ 'unique_product_id", "user_id", /* other fields */ ]模型上放置一个guarantee_tickets，因为除非您将可填写字段设置到模型中，否则它可能无法将字段保存在数据库中。