从 Laravel 5.1 中的通用数据库查询中获取 Eloquent 模型的实例

Get instance of Eloquent model from generic DB query in Laravel 5.1

我的模型设置了不同的关系。假设我的 Entry 模型属于 Supplier,所以通常我的模型文件中有一个 supplier() 方法。

到目前为止一切顺利,当我有类似 Entry::find(1)->supplier 的东西时,它的效果非常好。但是,当我从 Laravel 中的通用 DB:: 查询中获取条目时,不起作用的是,我显然无法访问 supplier() 方法,因为它不是 Entry.[= 的实例26=]

$entries = DB::table('suppliers')
            ->join('entries', "supplier.id", '=', "entries.supplier_id")
            ->select('entries.*')
            ->where("supplier.name", 'like', "%{$name}%")
            ->get();

现在如果我dd($entries)

我得到了预期的结果。但是当我做类似的事情时:

dd($entries[0]->supplier); // or ->supplier()

我收到这个错误:

Undefined property: stdClass::$supplier.

那么我如何 cast (?) 这些结果到 Entry Eloquent 模型以便我可以利用这些关系?


这是 $entriesprintr

Array
(
    [0] => stdClass Object
        (
            [id] => 1
            [user_id] => 0
            [archived] => 0
            [supplier_id] => 5
            [customer_id] => 1
            [contact] => dfgfdg
            [commission] => dfgdfg
            [entrance_date] => 2015-09-22 16:52:33
            [cost_estimate] => 1
            [status] => 1
            [type] => 1
            [watch_id] => 7
            [reference] => dfgdfg
            [serial_number] => 0
            [delivery_date] => 2015-09-07 16:52:33
            [articles_json] => 
            [total_sales_cost_netto] => 
            [gross_profit_netto] => 
            [gross_profit_brutto] => 
            [created_at] => 2015-09-09 20:10:02
            [updated_at] => 2015-09-11 16:52:33
        )

)

如@Zakaria 所述,只需使用 Eloquent:

$entries = Entry::with('supplier')
        ->join('supplier', "supplier.id", '=', "entries.supplier_id")
        ->where("supplier.name", 'like', "%{$name}%")
        ->get();

如果您确实需要 "cast" 它们,请尝试这样的操作:

$entries = $yourDbQuery;
$c = new \Illuminate\Database\Eloquent\Collection;
foreach ($entries as $entry) {
    $entryModel = new \App\Entry;
    $c->add($entryModel->forceFill((array)$entry));
}
$c->load('supplier');