問題描述

我需要使用 Laravel 進行原始數據庫查詢:

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");

我得到一個普通的 PHP StdClass 對象，其中包含成員表上的屬性字段.我想將其轉換為 Member(一個 Eloquent 模型實例)，如下所示:

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 IlluminateDatabaseEloquentModel;

class Member extends Model {
}

我不知道該怎么做，因為會員沒有設置任何字段，而且我擔心我不會正確初始化它.實現這一目標的最佳方法是什么?

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?

推薦答案

您可以嘗試將結果混合到模型對象中:

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() );

或者你甚至可以讓 Laravel 從原始查詢中為你自動補水:

Or you can even let Laravel auto-hydrate them for you from the raw query:

$models = Member::hydrateRaw( "SELECT * FROM members...");

編輯

從 Laravel 5.4 hydrRaw 開始不再可用.我們可以使用 fromQuery 代替:

From Laravel 5.4 hydrateRaw is no more available. We can use fromQuery instead:

$models = Member::fromQuery( "SELECT * FROM members..."); 

這篇關于如何從原始對象創建 Eloquent 模型實例?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！