我有不同关系的模型.假设我的Entry模型属于供应商,所以通常我的模型文件中有一个supplier()方法.
到目前为止一切都那么好,当我有一些像Entry :: find(1) – >供应商这样的产品时,效果非常好.然而,什么是无效的是当我从Laravel中的通用DB ::查询中获取条目时,我显然无法访问supplier()方法,因为它不是Entry的实例.
$entries = DB::table('suppliers')
->join('entries', "supplier.id", '=', "entries.supplier_id")
->select('entries.*')
->where("supplier.name", 'like', "%{$name}%")
->get();
现在,如果我dd($entries);
我得到了预期的结果.但当我做类似的事情时:
dd($entries[0]->supplier); // or ->supplier()
我收到此错误:
Undefined property: stdClass::$supplier.
那么如何将(?)这些结果转换为Entry Eloquent模型,以便我可以利用这些关系?
这是$条目的原则:
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
)
)
最佳答案 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();
如果你真的需要“施放”他们,尝试这样的事情:
$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');