Question

我是Laravel的初学者。我在我的项目Laravel 5.8中使用。

我有此代码：

$product = new Product();
$product->create([
    'company_id' => $request->user()->company_id,
    'enable' => $request->input('enable') ?? 0,
    'id_delivery_vat' => $request->input('id_delivery_vat') ?? 1,
    'id_product_vat' => $request->input('id_product_vat') ?? 1,
    'id_producer' => $request->input('id_producer'),
    'promo' => $request->input('promo') ?? 0,
    'name' => $request->input('name'),
    'qr_code' => $request->input('qr_code'),
    'oe_code' => $request->input('oe_code'),
    'description' => $request->input('description'),
    'product_price' => str_replace(",", ".", $request->input('product_price')) ?? 0,
    'promo_product_price' => str_replace(",", ".", $request->input('promo_product_price')) ?? 0,
    'product_delivery_price' => str_replace(",", ".", $request->input('product_delivery_price')) ?? 0,
    'quantity' => $request->input('quantity'),
    'url_address' => Str::slug($request->input('name'), '-')
]);

如何获取上一个插入ID？

我的产品迁移：

Schema::create('products', function (Blueprint $table) {
            $table->bigIncrements('id');
            $table->bigInteger('company_id')->unsigned();
            $table->foreign('company_id')->references('id')->on('companies')->onDelete('cascade');
            //$table->bigInteger('category_id')->unsigned();
            //$table->foreign('category_id')->references('id')->on('product_categories');
            $table->smallInteger('id_delivery_vat')->unsigned();
            $table->foreign('id_delivery_vat')->references('id')->on('vat');
            $table->smallInteger('id_product_vat')->unsigned();
            $table->foreign('id_product_vat')->references('id')->on('vat');
            $table->bigInteger('id_producer')->unsigned();
            //$table->foreign('id_producer')->references('id')->on('product_producers');
            $table->string('name', 120)->nullable();
            $table->string('qr_code', 120)->nullable();
            $table->string('oe_code', 120)->nullable();
            $table->char('enable', 1)->default(0);
            $table->char('promo', 1)->default(0);
            $table->longText('description')->nullable();
            $table->decimal('product_price', 9, 2)->default(0);
            $table->decimal('promo_product_price', 9, 2)->default(0);
            $table->decimal('product_delivery_price', 9, 2)->default(0);
            $table->unsignedInteger('quantity')->default(0);
            $table->string('url_address', 160);
            $table->timestamps();
            $table->engine = "InnoDB";
            $table->charset = 'utf8mb4';
            $table->collation = 'utf8mb4_unicode_ci';
        });

我的ID有bigIncrements。

型号：

class Product extends Model
{
    use scopeActiveTrait;

    protected $fillable = ['company_id', 'id_delivery_vat', 'id_product_vat', 'id_producer', 'name', 'qr_code', 'oe_code', 'enable', 'promo', 'description', 'product_price', 'promo_product_price', 'product_delivery_price', 'quantity', 'url_address'];
    protected $quarded = ['id'];
    public $timestamps = true;

    public function categories()
    {
        return $this->belongsToMany(Category::class, 'product_selected_categories', 'product_id', 'category_id');
    }
}

我该如何修理？我尝试：$ product-> id，$ product-> nextId（）-但这不起作用：（

Answer 1

$product->id是最后的插入ID

Answer 2

如果您希望将最后一个插入ID放入下一个插入产品中，则不必使用自动递增ID。

否则，如果您想将其用于其他用途，则只需保存产品，然后获取ID即可：

$product->save();
$lastId = $product->id; // or whatever name you give to the id

获取最后插入的ID记录到Laravel中的数据库

2 个答案: