I am using the Auto_Modeler class for data access and due to the nature of my app. I would like to use ` public static function check_AMexists($fieldname,$fieldvars,$model)
{
$thismodel = new Model_Admin_$model();`
It used to work with ORM where I would simply pass the $model to ORM:factory($model).
How do I go about passing the $model variable containing the modelname in my function?
$model_name = 'Model_Admin_' . $model;
$model = new $model_name;
Related
previously i made a navigation bar in a view page for which i wrote the function in corresponding controller but now i have to put this in main file (layout). But now i don't know where to write the function.
also i have to pass two variable to the layout file through the function that will contain the value of navigation bar.
I have already tried some method but it allows me to return only on value.
Basically i want to know that where can i write the below function to use it in main layout of yii2 basic
public function actionMenutest()
{
$query = new Query;
$data= $query->select('name,id')
->from('menu')->all();
$query2 = new Query;
$data2= $query2->select('name,menu_id')
->from('submenu')->all();
return $this->render('menutest',[
'data'=>$data, 'data2'=>$data2
]);
}
You can use EVENT_BEFORE_RENDER for this purpose. For advanced app the below code need to go into common\config\bootstrap.php file.
use yii\base\Event;
use yii\base\View;
Event::on(View::className(), View::EVENT_BEFORE_RENDER, function() {
$query = new Query;
$data= $query->select('name,id')
->from('menu')->all();
$query2 = new Query;
$data2= $query2->select('name,menu_id')
->from('submenu')->all();
Yii::$app->view->params['data'] = $data;
Yii::$app->view->params['data2'] = $data2;
});
Then in your main layout you can use your model as:
$data= $this->params['data'];
$data2= $this->params['data2'];
I have not used basic template as yet. But you can try the following:
Create a bootstrap.php file in config folder.
After that update the web/index.php file. Put the below code in that:
require(__DIR__ . '/../config/bootstrap.php');
Then put the above code in bootstrap.php file. Try it and let me know if you need any more help.
I am trying to paginate some categories and this error popped up from nowhere and I don't know how to fix it :
Call to a member function render() on a non-object
This is my controller:
public function getFilmeByCateg()
{
$categorii = Categorii::all();
//$Movies = Movies::all();
$categorie = Request::segment(2);
$cat = Categorii::where('denumire', '=',$categorie)->first();
$cat2 = Categorii_filme::where('categorie_id', '=' ,$cat->categorie_id)->get();
$filme = array();
foreach($cat2 as $filmulet)
{
$film = Movies::where('movie_id','=',$filmulet->film_id)->paginate(12)->first();
$filme[] = $film;
}
return view('filme')->with('Movies',$filme)->with('categorii',$categorii);
}
And this is how I render the paginator in my layout:
{!!$Movies->render() !!}
This is in my routes:
Route::get('categorie/{categorie}','WelcomeController#getFilmeByCateg');
This is in my Movies.php:
class Movies extends Model {
protected $table = "movies";
}
Can someone tell me how can I manage to make this work ?
Since $filme is an array, you can access render() method whose object you wanted is $Movies which is array.
foreach($cat2 as $filmulet)
{
$film = Movies::where('movie_id','=',$filmulet->film_id)->paginate(12)->first();
$filme[] = $film;
}
Here $filme is an array.
return view('filme')->with('Movies',$filme)->with('categorii',$categorii);
Here you passed $filme as $Movie variable.
{!!$Movies->render() !!}
And here you wanted to access render() method of $filme which is not an object.
Change your controller to this:
public function getFilmeByCateg()
{
$categorii = Categorii::all();
//$Movies = Movies::all();
$categorie = Request::segment(2);
$cat = Categorii::where('denumire', '=',$categorie)->first();
$movie_ids = Categorii_filme::where('categorie_id', '=' ,$cat->categorie_id)->get()->lists('movie_id');
$filme = Movies::whereIn('movie_id','=',$movie_ids)->paginate(12);
return view('filme')->with('Movies',$filme)->with('categorii',$categorii);
}
OK First of all, you made it the wrong way.
Using an intermediate class for representing the pivot table is completely bad.
I'll make it work for you.
Your Category class
class Category extends \Model {
protected $table = 'categories';
public function movies()
{
return $this->belongsToMany('Your\Namespace\To\Movie');
}
}
Your Movie class
class Movie extends \Model {
protected $table = 'movies';
public function categories()
{
return $this->belongsToMany('Your\Namespace\To\Category');
}
}
Your route is still :
Route::get('categorie/{category_id}','WelcomeController#getFilmeByCateg');
So in your controller method you do the following :
public function getFilmeByCateg($category_id)
{
$category = Category::findOrFail($category_id);
return view('filme')->with('Movies',$category->movies()->paginate(12)->ge())->with('categorii',Category::all());
}
Now in your view $Movies->render() will work
If you wont paginate your data how would you render it ? Documents on the official laravel website states :
There are several ways to paginate items. The simplest is by using the paginate method on the query builder or an Eloquent model.
Paging database results
$users = DB::table('users')->paginate(15);
Note: Currently, pagination operations that use a groupBy statement cannot be executed efficiently by Laravel. If you need to use a groupBy with a paginated result set, it is recommended that you query the database and create a paginator manually.
Creating A Paginator Manually
Sometimes you may wish to create a pagination instance manually,
passing it an array of items. You may do so by creating either an
Illuminate\Pagination\Paginator or
Illuminate\Pagination\LengthAwarePaginator instance, depending on your
needs.
Now in your case you should first
use Illuminate\Pagination\LengthAwarePaginator as Paginator;
and then
$paginator = new Paginator($items, $count, $limit, $page, [
'path' => $this->request->url(),
'query' => $this->request->query(),
]);
Please supply parameters accordingly in the Paginator function above.
I know it is very late but it will help someone who get in to same issue. Instead of ->get() use ->paginate(10) in query.
i have create an sql function in my database that take to Date params and get data from 5 tables.
after that add it to project as entity framework from database and the code generated is:
[DbFunction("Dr_EmploEntities", "SelectEmployee")]
public virtual IQueryable SelectEmployee(Nullable frm_date, Nullable to_date)
{
var frm_dateParameter = frm_date.HasValue ?
new ObjectParameter("frm_date", frm_date) :
new ObjectParameter("frm_date", typeof(DateTime));
var to_dateParameter = to_date.HasValue ?
new ObjectParameter("to_date", to_date) :
new ObjectParameter("to_date", typeof(DateTime));
return ((IObjectContextAdapter)this).ObjectContext.CreateQuery("[Dr_EmploEntities].[SelectEmployee](#frm_date, #to_date)", frm_dateParameter, to_dateParameter);
}
public DbSet SelectEmployee_Result { get; set; }
as you see i have now "SelectEmployee_Result" that don't take any params, and "SelectEmployee" that take two date params.
after that i have create an controller for "SelectEmployee_Result" class.
after that i run my project Index View that working with "SelectEmployee_Result" class give me err:
"The type 'SelectEmployee_Result' is mapped as a complex type. The Set method, DbSet objects, and DbEntityEntry objects can only be used with entity types, not complex types."
and i make breakpoint and see that "SelectEmployee_Result" has no data so i change the Index Code in controller and fill "SelectEmployee" with two date params
and when run got same err msg too.
so how can i fill "SelectEmployee_Result" from the beginning with data between two dates to let me use it in all views ?
all what i need here is view data i got i edit before saving it in database Like using DataTable but i need to do that from Entity with sql function
and what is difference between "SelectEmployee" that is my function name and that is need two params and "SelectEmployee_Result"?
I have created a View using module, now in controller of this view i need to fetch some specific content type and return to view. Please can some one eleborate with code sample.
You will need to inject the IContentManager services in your controller constructor (see dependency injection) , but since you will need to populate a new shape, you could inject IOrchardServices which will include a few common OrchardServices in one instance.
IOrchardServices services;
public MyController(IOrchardServices services){
this.services = services;
}
Then in your action (if you want to show it on the front end you will have to mark it as themed), do something like this:
[Themed]
public ActionResult MyAction(){
//Notice that you can filter the contentItems here, this is just a basic example
var myContentItems = services.ContentManager.Query().ForType("MyContentItem").List();
//You probably need to create a new shape for showing the ContentTypes
var shape = services.New.YourCustomShape(); //Notice that you must create a view that matches this name
shape.YourContentItems = myContentItems;
return new ShapeResult(this, shape);
}
And that's it.
I have a product entity and it has an images field that store the images names from the product but the images names depends of a part_number field that is unique, so if the user make a mistake in the part number and he wants to edit it then I also have to change the images names
I tried this but it does not works:
// class ProductsAdmin extends Admin
public function preUpdate($product) {
$old_product = $this->getSubject();
if ($old_product->getPartNumber() != $product->getPartNumber)
{
// change file names
}
$this->saveFile($product);
}
How I get the original row in preUpdate() function?
According to the topic taken from the official SonataAdmin google forum:
https://groups.google.com/forum/#!topic/sonata-devs/0zML6N13i3U
you need to make use of the class UnitOfWork:
http://www.doctrine-project.org/api/orm/2.3/class-Doctrine.ORM.UnitOfWork.html
Do this way:
public function preUpdate($object)
{
$em = $this->getModelManager()->getEntityManager($this->getClass());
$original = $em->getUnitOfWork()->getOriginalDocumentData($object);
}
Thus you get an array of values of your database entity.
E.g: to get access to the value password of your entity do:
$password = $original['password'];
That's all.
Enjoy :)
If you just do a doctrine query in the preUpdate function to get the product from the database you'll have the old object. Then do the comparison and you're good to go.