Laravel：通过使用“ With”，是否可以避免对数据库的多次查询/连接？

以下面的数据库结构为例：

Table Sales
id | customer_id | currency_id | number | due_date

Table Open Accounts
id | sale_id | credit | debit | balance

现在说我要获得所有未结帐户：

SELECT OpenAccounts::select('oa.*')
    ->from('open_accounts As oa')
    ->join('sales AS sale', 'sale.id', '=', 'oa.sale_id')
    ->where('oa.balance', '>', 0)
    ->orderBy('sale.due_date', 'ASC')
    ->get();

如果要遍历所有结果并打印出一些数据，可以执行以下操作：

@foreach ($openAccounts as $open)
    <tr>
        <td>{{ $open->Sale->due_date }}</td>
        <td>{{ $open->Sale->Customer->name }}</td>
        <td>{{ sprintf('%s %s', $open->balance, $open->Sale->Currency->symbol) }}</td>
    </tr>
@endforeach

问题：这意味着对于每个结果，我都打开一个MySQL连接并查询Sale表，Customer表和Currency表？

现在，如果我将代码更改为：

SELECT OpenAccounts::select('oa.*')
    ->with('Sale')
    ->with('Sale.Customer')
    ->with('Sale.Currency')
    ->from('open_accounts As oa')
    ->join('sales AS sale', 'sale.id', '=', 'oa.sale_id')
    ->where('oa.balance', '>', 0)
    ->orderBy('sale.due_date', 'ASC')
    ->get();

通过打印输出，我得到如下信息：

Illuminate\Database\Eloquent\Collection Object
(
    [0] => OpenAccounts Object
    (
        [attributes:protected] => Array
        (
            [id] => 89
            [sale_id] => 150
            [debit] => 184.93
            [credit] => 0
            [balance] => 184.93
        )
        [relations:protected] => Array
        (
            [Sale] => Sales Object 
            (
                [attributes:protected] => Array
                (
                    [id] => 150
                    [customer_id] => 1
                    [currency_id] => 1
                    [number] => 5
                    [due_date] => '2020-03-08'
                )
                [relations:protected] => Array
                (
                    [Customer] => ...
                    [Currency] => ...
                )
            )
        )
    )
)

问题：这是否意味着现在在$open->Sale->Customer->name循环中使用示例时，我没有初始化新的MySQL连接，而是直接访问Eloquent Collection返回的属性？