laravel：模型中的函数必须返回一个关系实例
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laravel: function in model must return a relationship instance
提问by Ali
I try to build a path for a model on laravel
我尝试在 Laravel 上为模型构建路径
I created a function in my model:
我在我的模型中创建了一个函数：
public function path()
{
return App\Helper\GeneralController::getURL($this);
}
with dd(App\Helper\GeneralController::getURL($this))test I got the right answer. (output is a URL)
通过dd(App\Helper\GeneralController::getURL($this))测试，我得到了正确的答案。（输出是一个 URL）
but in view with the call: $article->pathI get this error:
但鉴于电话：$article->path我收到此错误：
App\Article:: path must return a relationship instance.
App\Article:: path must return a relationship instance.
What is wrong?
怎么了？
回答by Alexey Mezenin
You need to call it:
你需要调用它：
$article->path()
When you do $article->path, you're trying to use Eloquent relationship which you don't have.
当您这样做时$article->path，您正在尝试使用您没有的 Eloquent 关系。
回答by Ali
I faced that error when I forgot to write returnbefore relation in the model!
check it out now!
当我忘记在模型中的关系之前写返回时，我遇到了那个错误！
现在就去看看！
回答by pwyg
I know this has already been answered and accepted. However, if the OP did want to use a property accessor rather than a method use the "get{property name}Attribute" syntax of Laravel to create a custom attribute.
我知道这已经得到了回答和接受。但是，如果 OP 确实想使用属性访问器而不是方法，请使用 Laravel 的“get{property name}Attribute”语法来创建自定义属性。
Here is what it would look like for this specific case:
下面是这个特定案例的样子：
public function getPathAttribute()
{
return App\Helper\GeneralController::getURL($this);
}
using this approach "path" can now be called as an attribute and will not be resolved to a relationship using the syntax:
使用这种方法，“路径”现在可以作为属性调用，并且不会使用以下语法解析为关系：
$article->path;
回答by jynk
You're calling a relationship.
你称之为关系。
$article->path
To call the method, use '()', like so,
要调用该方法，请使用“()”，如下所示，
$article->path()
回答by Niklesh Raut
path()is method not object element you need to call as method
path()是方法不是您需要作为方法调用的对象元素
$article->path();
