类型提示雄辩模型 [英] Type Hinting Eloquent Models

问题描述

我正在尝试编写好的代码，其中一部分是类型暗示，以便使工作更容易，并强制期望。

I am trying to write good code, and part of that is type hinting to make it easier for work down the line and to force expectations.

这可能看起来有点不舒服，但更多的是我的概念证明。

This may seem a little contrived, but its more of a proof of concept for me.

我是编写一个类，将一个TSV文件拆分在标签上并插入到我的模型中。在我的构造函数中，我要求：

I am writing a class to take a TSV file split on tabs and insert into my Model. In my constructor I was asking for:

Illuminate\Database\Eloquent\Model

我通过了：

new \App\Model()

最后的错误响应：

instance of App\Model given

显然我做错了，但是我不想强制使用App \Model，我如何一般地要求一个有说服力的模型？

Clearly I have done something wrong, but I do not want to force usage of App\Model, how can I generically ask for an eloquent model?

编辑更多信息

为了使之更清楚，我使用的是Laravel 5，这些模型是通过artisan make：model创建的。构造函数如下：

To make it more clear, I am using Laravel 5, the models are created via artisan make:model. The constructor is as follows:

function __construct ($resource, Illuminate\Database\Eloquent\Model $model, $skip = 0)

而我正在使用的模型（对于我的电影表）是：

And the Model I am using (for my movie table) is:

use Illuminate\Database\Eloquent\Model;

class Movie extends Model {

推荐答案

p>在您的类型提示中，使用反斜杠前缀FQCN：

In your type hint, preface the FQCN with a backslash:

function __construct ($resource, \Illuminate\Database\Eloquent\Model $model, $skip = 0)

或者， code>使用语句到你的班级：

Either that, or add the use statement to your class:

use Illuminate\Database\Eloquent\Model;

class MyClass {
    function __construct ($resource, Model $model, $skip = 0) {
        //
    }
}

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