从雄辩的集合中获取对象的嵌套值

Question

我正在将laravel的收银员与stripe＆一起从Invoices集合中使用，并且我正尝试通过其subscription密钥检索特定的发票：

Collection {#157 ▼
  #items: array:1 [▼
    0 => Invoice {#162 ▼
      #owner: User {#163 ▶}
      #invoice: Invoice {#208 ▼
        +"id": "in_1BsZF2F3eLup6dYnvs74ij6B"
        +"object": "invoice"
        ...
        +"statement_descriptor": null
        +"subscription": "sub_CH7UxbgcTCeHLp"
        +"subtotal": 1999
        +"tax": null
        +"tax_percent": null
        +"total": 1999
        +"webhooks_delivered_at": 1517934260
      }
    }
  ]
}

这是user模型用于检索此Invoices集合的函数：

public function invoices($includePending = false, $parameters = [])
{
    $invoices = [];

    $parameters = array_merge(['limit' => 24], $parameters);

    $stripeInvoices = $this->asStripeCustomer()->invoices($parameters);

    // Here we will loop through the Stripe invoices and create our own custom Invoice
    // instances that have more helper methods and are generally more convenient to
    // work with than the plain Stripe objects are. Then, we'll return the array.
    if (! is_null($stripeInvoices)) {
        foreach ($stripeInvoices->data as $invoice) {
            if ($invoice->paid || $includePending) {
                $invoices[] = new Invoice($this, $invoice);
            }
        }
    }

    return new Collection($invoices);
}

每个Invoice对象都有两个属性， Owner （当前用户）和实际Invoice

向下钻取每个Invoices subscription密钥并对照该值进行核对的正确语法是什么？

我尝试了以下查询，结果为NULL ：

$user->invoices()->where('subscription','sub_CH7UxbgcTCeHLp')->first();

更新

当我转储以下查询$user->invoices()->first()->subscription它在sub_CH7UxbgcTCeHLp返回正确的字符串，但是当我尝试使用$user->invoices()->firstWhere('subscription', 'sub_CH7UxbgcTCeHLp')其结果为NULL

Answer 1

请尝试以下操作：

$id = 'sub_CH7UxbgcTCeHLp'
$user->invoices->first(function ($invoice) use ($id) {
    return $invoice->invoice->subscription === $id;
});

尝试使用where时，似乎只是缺少实际发票对象上的嵌套发票属性。 这可能起作用：

$id = 'sub_CH7UxbgcTCeHLp';
$user->invoices()->where(function ($query) use ($id) {
    $query->where('invoice.subscription', $id);
})->first();