将Laravel的查询结果上传到课堂 [英] Cast Laravel query results to class

问题描述

使用语法 DB :: table（'foo'）创建查询时，会创建一个泛型类（stdClass）。有没有办法将生成的行转换到特定的类？

When creating a query using the syntax DB::table('foo'), it creates a generic class (stdClass). Is there any way to cast the resulting rows to a specific class?

这里是一些示例代码，应该解释我想做什么：

Here is some example code that should explain what I want to do:

$result = DB::table('foo')->get();
$converted = (Foo) $result; // Not going to work

我想将数组的所有（stdClass）对象转换为Foo类。

I want to cast all the (stdClass) objects of the array to the Foo class.

推荐答案

是的，您可以将结果合并到所需的类中。我发现这个答案埋在一堆混乱的半答案和困惑的问题，弥补了可怕的 Laracasts.com论坛。感谢您在这里提问，而不是在那里。

Yes, you can hydrate the results into the classes you want. I found the answer buried deep in a mess of half-answers and confused questions that make up the terrible Laracasts.com forum. Thanks for asking the question here instead of there.

一旦你得到结果，就可以使用你的模型类来达到这个目的：

Once you get the results, hydrate them using your model class:

$result = DB::table('foo')->get();
$converted = Foo::hydrate($result);

编辑：找到一些文件在水合物方法也如此这有助于

Found some documentation on the hydrate method too, if that helps

编辑2：我发现自己处于需要从数组或集合，具体取决于查询的结果。当一个集合被返回时，它被正确地水化，但是当结果是数组时，它们只是 stdClass 。 我写了一个快速方法添加到我的主人模型采集数组或对象或分页对象的集合，并将其正确地转换为我想要的对象。

Edit 2: I found myself in a situation where I needed to cast results from either an array or a collection, depending on the results of a query. When a collection was returned, it was correctly hydrated, but when the result was an array, they were just stdClass. I wrote a quick method added to my master model that took a collection of arrays or objects, or a pagination object, and correctly cast it to the object I wanted.

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