php - 在不检索实例的情况下获取相关模型(Laravel 4，Eloquent)的类(名称)

Question

$relationFunctionName = 'bananas';    //this is set dynamically at runtime, and is always a relation function
$currentClass         = 'FruitBowl'; //this is set dynamically at runtime, and is always an Eloquent Model
$rowId                = 1; //this is also set dynamically at runtime

$grabber              = new $currentClass();
$item                 = $grabber->with($relationFunctionName)->find($rowId);
$relatedItem          = $item->{$relationFunctionName};

exit( get_class( $relatedItem ) );
// returns the className of the related item (if there is one) returns the current class if there is none

我还研究了“getRelations()”，它返回一个数组，其中关系函数名称作为填充相关项的键，如果没有则返回 NULL。

我也可以创建一个新的关系实例，这样我就可以检索它的类名而不是保存它。

明确地说，我想要关系函数返回的对象的类名。所以在这个例子中它可能是香蕉。

Answer 1

好的，找到了。结果很简单。从上面的代码继续，执行以下操作将为您提供类名:

$relationFunctionName = 'bananas';    //this is set dynamically at runtime, and is always a relation function
$currentClass         = 'FruitBowl'; //this is set dynamically at runtime, and is always an Eloquent Model
$rowId                = 1; //this is also set dynamically at runtime

$grabber              = new $currentClass();
$modelName            = $grabber->{$relationFunctionName}()->getModel();

$modelName 现在充满了类本身..所以如果你想要 className 它应该是 get_class( $modelName );