Laravel 按表名获取模型
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Laravel get Model by Table Name
提问by Defy
Is there any way to get a model by table name?
有没有办法通过表名获取模型？
For example, I have a "User" model, its table is defined as protected $table = "users"
例如，我有一个“用户”模型，它的表定义为 protected $table = "users"
Now, what I want to do is to get the model by table name which is equal to "users".
现在，我想要做的是通过等于“users”的表名来获取模型。
This function is more like the reverse of Model::getTable();
这个功能更像是 Model::getTable();
I have searched everywhere but I could not find a solution, perhaps I might be missing something simple?
我到处搜索，但找不到解决方案，也许我可能遗漏了一些简单的东西？
EDIT
编辑
I am building something like an API :
我正在构建类似 API 的东西：
Route::get('/{table}', 'ApiController@api');
Route::get('/{table}/filter', 'ApiController@filter');
Route::get('/{table}/sort', 'ApiController@sort');
Route::get('/{table}/search', 'ApiController@search');
so in the address bar, for example when I search for the "users", I could just hit on the URL:
所以在地址栏中，例如当我搜索“用户”时，我可以直接点击 URL：
api/users/search?id=1
then on the controller, something like:
然后在控制器上，类似于：
public function search(){
// get all the params
// get the model function
$model = //function to get model by table name
// do some filtering, then return the model
return $model;
}
回答by Igor R
Maybe something like this will help you:
也许这样的事情会帮助你：
$className = 'App\' . studly_case(str_singular($tableName));
if(class_exists($className)) {
$model = new $className;
}
回答by huuuk
You must determine for which table name which class to call. I see 2 ways to do this.
您必须确定要为哪个表名调用哪个类。我看到有两种方法可以做到这一点。
Use Laravel's models naming convention as @IgorRynkovoy suggested
按照@IgorRynkovoy 的建议使用 Laravel 的模型命名约定
or
或者
Use some kind of dictionary
使用某种字典
public function search($tableName)
{
$dictionary = [
'table_name' => 'CLASS_NAME_WITH_NAMESPACE',
'another_table_name' => 'CLASS_NAME_WITH_NAMESPACE',
];
$className = $dictionary[$tableName];
$models = null;
if(class_exists($className)) {
$models = $className::all();
}
// do some filtering, then return the model
return $models;
}
回答by Mauro Baptista
I know that it is an old question, but it can help someone:
我知道这是一个老问题，但它可以帮助某人：
public function getModelFromTable($table)
{
foreach( get_declared_classes() as $class ) {
if( is_subclass_of( $class, 'Illuminate\Database\Eloquent\Model' ) ) {
$model = new $class;
if ($model->getTable() === $table)
return $class;
}
}
return false;
}
It will return the class name, so you need to instantiate it.
它将返回类名，因此您需要实例化它。
回答by Oleg
Alternative variant. I have my base model App\Models\Model This model have static method getModelByTable, ofcourse you can store this method anywhere you want.
替代变体。我有我的基本模型 App\Models\Model 这个模型有静态方法 getModelByTable，当然你可以把这个方法存储在你想要的任何地方。
public static function getModelByTable($table)
{
if (!$table) return false;
$model = false;
switch ($table) {
case 'faq':
$model = Faq::class;
break;
case 'faq_items':
$model = FaqItems::class;
break;
}
if ($model) {
try {
$model = app()->make($model);
} catch (\Exception $e) {
}
}
return $model;
}
回答by Marcus Christiansen
studly_case()and str_singular()are deprecated functions.
studly_case()并且str_singular()是不推荐使用的函数。
You can use the Illuminate\Support\Strfacade.
您可以使用Illuminate\Support\Str门面。
$className = 'App\' . Str::studly(Str::singular($tableName));
