如何从原始对象创建一个雄辩的模型实例? [英] How to create a Eloquent model instance from a raw Object?
问题描述
我需要使用Laravel进行原始数据库查询:
I need to make a raw database query using Laravel:
$results = DB::select("SELECT * FROM members
INNER JOIN (several other tables)
WHERE (horribly complicated thing)
LIMIT 1");
我返回一个简单的PHP StdClass对象,其中包含用于成员表的属性的字段.我想将其转换为如下所示的Member(一个雄辩的模型实例):
I get back a plain PHP StdClass Object with fields for the properties on the members table. I'd like to convert that to a Member (an Eloquent model instance), which looks like this:
use Illuminate\Database\Eloquent\Model;
class Member extends Model {
}
由于成员没有设置任何字段,我不确定如何执行此操作,并且我担心无法正确初始化它.最好的方法是什么?
I'm not sure how to do it since a Member doesn't have any fields set on it, and I'm worried I will not initialize it properly. What is the best way to achieve that?
推荐答案
您可以尝试将结果合并到Model对象中:
You can try to hydrate your results to Model objects:
$results = DB::select("SELECT * FROM members
INNER JOIN (several other tables)
WHERE (horribly complicated thing)
LIMIT 1");
$models = Member::hydrate( $results->toArray() );
或者您甚至可以让Laravel从原始查询中自动为它们补水:
Or you can even let Laravel auto-hydrate them for you from the raw query:
$models = Member::hydrateRaw( "SELECT * FROM members...");
编辑
从Laravel 5.4开始, hydrateRaw 不再可用.我们可以改用 fromQuery :
From Laravel 5.4 hydrateRaw is no more available. We can use fromQuery instead:
$models = Member::fromQuery( "SELECT * FROM members...");
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