Yii calling controller function from another controller - php

Question

In my web application the work flow demands that I should call one controller function from another function
Should I do add extra code or some configuration to proceed ? Right now This is how I implemented .But When I click "save" button nothing is happening values are just getting empty from the form .
My code .I want to create an object of model "BookVegetable" inside "ProducerOfferController" .
My code inside producerOffer controller
public function actionCreate()
{
//$book_vegetable=new BookVegetable;
$model=new BookVegetable;
if(isset($_POST['BookVegetable']))
{
$model->attributes=$_POST['BookVegetable'];
$model->booked_by = Yii::app()->user->id;
$model->save();
if ($model->hasErrors() === false)
{
$this->redirect(Yii::app()->user->returnUrl);
}
}
else
{
Yii::app()->user->setReturnUrl($_GET['returnUrl']);
}
$this->render('book',array('model'=>$model,));
}
My code for from view
<div style='padding-left:50px'>
<?php $form=$this->beginWidget('bootstrap.widgets.TbActiveForm',array('id'=>'non-ajax_form','enableAjaxValidation'=>false,)); ?>
<p class="help-block">Fields with <span class="required">*</span> are required.</p>
<?php echo $form->errorSummary($model); ?>
<?php echo "<br>" ?>
<?php echo CHtml::textField("booked_quantity",$model->booked_quantity); ?>
My scenario
public function actionBookvegetable($id){
$BookVegetable=new BookVegetable;
$model=$this->loadModel($id);
if(isset($_POST['ProducerOffer'],$_POST['BookVegetable']))
{
$model->attributes=$_POST['ProducerOffer'];
$BookVegetable->attributes=$_POST['BookVegetable'];
$BookVegetable->booked_by=Yii::app()->user->id;
$BookVegetable->producer_offer_id=$model->id;
$model->save();
$BookVegetable->save();
if (($model->hasErrors() === false)||($BookVegetable->hasErrors()=== false))
{
$this->redirect(Yii::app()->user->returnUrl);
}
}
else
{
Yii::app()->user->setReturnUrl($_GET['returnUrl']);
}
$this->render('book',array('model'=>$model,'BookVegetable'=>$BookVegetable));
}
<div class="form-actions">
<?php $this->widget('bootstrap.widgets.TbButton', array('buttonType'=>'submit', 'type'=>'primary', 'label'=> 'Save',)); ?>
How should I resolve this ? Is it essential to add anything extra to use one controller action inside another controller
The url before saving and its the same after I press save also
http://localhost/xxx/producerOffer/bookvegetable/20?returnUrl=%xxx%2FproducerOffer%2Fmanage

Answer 1

One way to do this in Yii2
Background
In SiteController action index method get the records of all objects from another model called voyzes.
In SiteController
Include the other model ex. Voyzes
Now in the SiteController action index method, implement the code to access the model/SQL/NoSQL or anything and set it in a array and return it to the view. For ex.
Now in the index view you should have your data from another model.

Answer 2

According to your information provided.. when you go here,
http://xxx.yyy.zzz/xxxx/producerOffer/create
Actually it should show you the form of book and when you click the save button there and go to returnUrl.
$this->redirect(Yii::app()->user->returnUrl);
I suggest you to write the following as,
$model->save();
$BookVegetable->save();
if (($model->hasErrors() === false)||($BookVegetable->hasErrors()=== false))
{
$this->redirect(Yii::app()->user->returnUrl);
}
To
if($model->save() && $BookVegetable->save())
$this->redirect('yourAction'); //if params needed, $this->redirect(array('yourAction', 'id' => $model->id));
When you go here, http://xxx.yyy.zzz/xxxx/producerOffer/bookvegetable
What ever the code you have written under ActionBookvegetable will trigger.
To make sure your values submitted properly please change this code
$model->save();
if ($model->hasErrors() === false)
{
$this->redirect(Yii::app()->user->returnUrl);
}
To
if($model->save())
$this->redirect('yourAction');
else
print_r(getErrors());
This will print any errors thats preventing from saving the model. let me know after you try this.

Answer 3

In my work , I have a lot of logic around the "loadModel" routine in controllers to make sure the user logged in has access to the particular model. I found this to work from another controller when I need to access the model, without moving or re-copying the loadmodel routine:
$caseviewController = Yii::app()->createController('Caseview');
//use this method from caseview controller to securley load case view model
$caseview = $caseviewController[0]->loadModel($caseviewid);