通过字符串调用 Laravel 模型
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Call Laravel model by string
提问by Parampal Pooni
Is it possible to call a Laravel model by string?
是否可以通过字符串调用 Laravel 模型？
This is what i'm trying to achieve but its failing:
这就是我想要实现的目标，但它失败了：
$model_name = 'User';
$model_name::where('id', $id)->first();
I get the following exception:
我收到以下异常：
exception 'ErrorException' with message 'Undefined variable: User'
回答by patricus
Yes, you can do this, but you need to use the fully qualified class name:
是的，您可以这样做，但您需要使用完全限定的类名：
$model_name = 'App\Model\User';
$model_name::where('id', $id)->first();
If your model name is stored in something other than a plain variable (e.g. a object attribute), you will need to use an intermediate variable in order to get this to work.
如果您的模型名称存储在普通变量（例如对象属性）以外的其他内容中，您将需要使用中间变量才能使其工作。
$model = $this->model_name;
$model::where('id', $id)->first();
回答by Emeka Mbah
Try this:
尝试这个：
$model_name = 'User';
$model = app("App\Model\{$model_name}");
$model->where('id', $id)->first();
回答by Zia
Yes you can do it with more flexible way. Create a common function.
是的，你可以用更灵活的方式来做到这一点。创建一个通用函数。
function getWhere($yourModel, $select, $is_model_full_path=null, $condition_arr=null)
{
if(!$is_model_full_path){ // if your model exist in App directory
$yourModel ="App\$yourModel";
}
if(!$condition_arr){
return $yourModel::all($select)->toArray();
}
return $App_models_yourModel::all($select)->where($condition_arr)->toArray();
}
Now you can call this method in different ways.
现在您可以通过不同的方式调用此方法。
getWhere('User', ['id', 'name']); // with default path of model
getWhere('App\Models\User', ['id', 'name'], true); // with custom path of model
getWhere('User', ['id', 'name'], false, ['id', 1]); // with condition array
By the way I like to use such functions.
顺便说一下，我喜欢使用这样的功能。
回答by Manojkiran.A
if you use it in many of the places use this function
如果您在许多地方使用它，请使用此功能
function convertVariableToModelName($modelName='',$nameSpace='App')
{
if (empty($nameSpace) || is_null($nameSpace) || $nameSpace === "")
{
$modelNameWithNameSpace = "App".'\'.$modelName;
return app($modelNameWithNameSpace);
}
if (is_array($nameSpace))
{
$nameSpace = implode('\', $nameSpace);
$modelNameWithNameSpace = $nameSpace.'\'.$modelName;
return app($modelNameWithNameSpace);
}elseif (!is_array($nameSpace))
{
$modelNameWithNameSpace = $nameSpace.'\'.$modelName;
return app($modelNameWithNameSpace);
}
}
Example if you want to get all the user
例如，如果您想获取所有用户
Scenario 1:
场景一：
$userModel= convertVariableToModelName('User');
$result = $userModel::all();
Scenario 2:
场景2：
if your model in in custom namespace may be App\Models
如果您的模型在自定义命名空间中可能是 App\Models
$userModel= convertVariableToModelName('User',['App','Models']);
$result = $userModel::all();
Hope it helps
希望能帮助到你
回答by itepifanio
This method should work well:
这种方法应该很有效：
private function getNamespace($model){
$dirs = glob('../app/Models/*/*');
return array_map(function ($dir) use ($model) {
if (strpos($dir, $model)) {
return ucfirst(str_replace(
'/',
'\',
str_replace(['../', '.php'], '', $dir)
));
}
}, array_filter($dirs, function ($dir) use ($model) {
return strpos($dir, $model);
}));
}
My project has multiple subdirectory and this function work well. So I recover the namespace of the model with $model = current($this->getNamespace($this->request->get('model')));Then I just have to call my query: $model::all()
我的项目有多个子目录，这个功能运行良好。所以我恢复了模型的命名空间$model = current($this->getNamespace($this->request->get('model')));然后我只需要调用我的查询：$model::all()
回答by Sheetal Mehra
You can also try like this:-
你也可以这样试试：-
use App\Model\User;
class UsersController extends Controller
{
protected $model;
public function __construct(User $model)
{
$this->model = $model;
}
public function getUser()
{
$data = $this->model->where('id', $id)->first();
return $data;
}
}
