I have an index action in one of my controllers.
In this index action i wish to create a "search" form that calls another action within my controller with a post request.
All the documentation i could find on form creation in cakephp is about creating new elements (i.e insert data into a database ) and not actually sending data to another action / function.
here is an example:
<?php echo $this->Form->create('Product'); ?>
<fieldset>
<legend><?php echo __('Søg Produkt'); ?></legend>
<?php
echo $this->Form->input('Search field');
?>
</fieldset>
<?php echo $this->Form->end(__('Søg')); ?>
How would i send the value of my search field to another action? (so it redirects to that action and sends data to it)
Did you try to read the documentation?
$this->Form->create('Product', array('url' => array('action' => 'search');
Related
I am a bit new to the Yii Framework. I am making a product selling website, which has 3 basic models
1. Users model containing the primary key id
2. Products model containing the primary key id
3. Orders model which is basically a mapping between the products and orders. It contains the fields product_id and user_id as foreign keys.
I have made a page where all the products are populated and the logged in user can click on a button on product box to order a particular product.
the code of the link is like this
<?php echo CHtml::link('Order Now',array('order',
'product_id'=>$model->id,
'user_id'=>Yii::app()->user->id)); ?>
(Q1) This is sending a GET request but I want to sent the details as post request. How to do this?
My default controller is the site controller. I have made an actionOrder method in this controller.
The code is:
if(Yii::app()->user->isGuest){
$this->redirect('login');
}else{
$model=new Orders;
if(isset($_POST['products_id']))
{
$model->attributes->products_id=$_POST['product_id'];
$model->attributes->users_id=Yii::app()->user->id;
if($model->save())
$this->redirect(array('index'));
}
$this->render('index');
}
But this code is showing bunch of errors. Also, (Q2) how can I put both products_id and users_id in a single array Orders so that I just have to write $_POST['orders']
Also, (Q3) how can I display a flash message after the save is successful?
Kindly help me to solve my 3 problems and sorry if you feel that the questions are too stupid.
Q1: If you want to use POST request, you're going to have to use a form of sorts, in this case the CActiveForm.
Controller:
public function actionOrder()
{
if(Yii::app()->user->isGuest)
$this->redirect('login');
else
{
$model=new Orders;
if(isset($_POST['Orders']))
{
$model->product_id=$_POST['Orders']['products_id'];
$model->users_id = Yii::app()->user->id;
if($model->save())
{
// Q3: set the flashmessage
Yii::app()->user->setFlash('ordered','The product has been ordered!');
$this->redirect(array('index'));
}
}
$this->render('index', array('model'=>$model)); //send the orders model to the view
}
}
View:
<!-- Q3: show the flash message if it's set -->
<?php if (Yii::app()->user->hasFlash('ordered')): ?>
<?php echo Yii::app()->user->getFlash('ordered'); ?>
<?php endif ?>
...
<?php $form=$this->beginWidget('CActiveForm', array('id'=>'order-form')); ?>
<?php echo $form->hiddenField($model,'products_id',array('value'=>$product->id)); ?> // please note the change of variable name
<?php echo CHtml::submitButton('Order Now'); ?>
<?php $this->endWidget(); ?>
Please note that I have changed the name of the product model variable $model to $product, because we will be using $model for the Orders model for the form.
Q2: In this case I set the users_id value in the controller, so $_POST['Orders'] only contains the value for products_id. In yii you can also mass assign your attributes with:
$model->attributes = $_POST['Orders']
Which basicly means $_POST['Orders'] is already an associative array containing the attribute names and values that are in your form.
Q3: The code shows you how to set and show a flash message after an order is succesfull.
First you have to declare forms send method, if you're using bootsrap it'll be like mine:
<?php $form = $this->beginWidget('bootstrap.widgets.TbActiveForm', array(
'action' => Yii::app()->createUrl($this->route),
'method' => 'post',
'id' => 'activity_timeRpt',
));
?>
Second if you want to send custom inputs, you have to specify, otherwise it'll be like
i'll be back to finish this
For your questions 1 and 2 I'd recomend you to use a CActiveForm class. For example
<?php $form = $this->beginWidget('CActiveForm', array(
'action' => 'you_action_here'
'method'=>'post' // this is optinal parameter, as 'post' is default value
)); ?>
<?php echo $form->textField($model,'product_id'); ?>
<?php echo $form->hiddenField($model,'user_id', array('value'=>Yii::app()->user->id)); ?>
<?php $this->endWidget(); ?>
where $model is instance of Orders class, passed by variables thru controller, or set in view file. After that you can use it in way you wanted $model->attributes = $_POST['orders'] in your action method.
For flash message you can use Yii->app()->user->setFlash('orderStatus', 'Successful'), before redirect( or render ) in your actionOrder. To show it:
<?php if(Yii::app()->user->hasFlash('orderStatus')):?>
<div class="info">
<?php echo Yii::app()->user->getFlash('orderStatus'); ?>
</div>
<?php endif; ?>
I have a little problem with a Controller method AJAX call in Yii. The thing is that I'm trying to filter the data of one dropDownList based in the value of a previous selected item.
In the view file, where I figured out is the source of the problem, I have this piece of code:
<?php echo $form->labelEx($model,'Estado'); ?>
<?php echo $form->dropDownList($model,'estado',CHtml::listData(Estado::model()->findAll(),'id','nombre'),array(
'ajax'=>array(
'type'=>'POST',
'url'=>CController::createAbsoluteUrl('buscar/select'),
'update'=>'#'.CHtml::activeId($model,'tbl_municipio_id'),
),
'class'=>'form-control'
));
?>
<?php echo $form->error($model,'Estado'); ?>
On the Controller side, I got this:
public function actionSelect(){
echo "Hello world";
$data = Municipio::model()->findAll('tbl_estado_id=:tbl_estado_id',
array(':tbl_estado_id'=>(int) $_POST['Consultorio_estado']));
$data = CHtml::listData($data,'id','name');
foreach($data as $value=>$name)
{
echo CHtml::tag('option',
array('value'=>$value),CHtml::encode($name),true);
}
The ajax call to the Select method isn't triggered when the dropDownList is clicked. I tracked the request using Firebug and no error nor fail message is dropped.
Anyone knows what can I do?.
Thanks in advance.
With my knowledge in Yii 1.1.13, there is no such option for ajax for form->dropDownList, just Chtml::dropDownList does.
Therefore you have option to manually custom event change of form->dropDownList or add more jQuery script to handle it by yourself, or simply switch to use Chtml::dropDownList like below example
<?php
echo CHtml::dropDownList('inst_province','',
array(1=>'A',2=>'B',3=>'C', 4=>'D'),
array(
'prompt'=>'Select City',
'ajax' => array(
'type'=>'POST',
'url'=>CController::createUrl('city/selectAll'),
'update'=>'#city_area',
'data'=>array('city_param'=>'js:this.value'),
)));
?>
http://www.yiiframework.com/wiki/429/an-easy-solution-for-dependent-dropdownlist-using-ajax/
I'm trying to merge 3 models to create a fourth one. I have model1, model2 and model3 and I want to merge them into modelMaster. I've also created controllers for all of them. When I call modelMaster/create action, I render the modelMaster/create view which renders the modelMaster/_form view. Inside this _form view, I also want to render model1/_form, model2/_form and a CHtml::dropDownList(), wich takes datas from model3. However, this doesn't work. How can I combine these three different views into one another?
If you try to skip the form generate from the _form views and use unique model names, I think you can use this manual: single form with more models
So the generate of the form definition handles always the parent view and the _form's only the inputs
The other way to use single model in views, create a form model by extend CFormModel, and handle the data binding between this model and the wrapped submodels
If you want to nest several forms into one master form you have to adjust the form templates accordingly. All of your modelMaster/create, model1/_form, model2/_form-views create and render a new CActiveForm (and thus several <form> tags).
Since you cannot nest form-elements in html (how should html know which action to pass the data to) you have to avoid this by doing the following:
Extract the inputs of the form you want to nest into a new view, e.g. model1/_formInputs would look like
...
<?php echo $form->labelEx($model,'name'); ?>
<?php echo $form->textField($model,'name');
<?php echo $form->error($model,'name');
...
alter the model1/create and the other views and get a reference to the form created there, by assigning the return of $this->beginWidget to the variable $form (if not already done):
<?php $form = $this->beginWidget('CActiveForm', array(
'id'=>'foo',
)); ?>
replace the former input fields with
<?php $this->renderPartial('model1/_formInputs', array('form' => $form); ?>
Now, for example the old model1/create-view should work as expected
To get your multi-model-form working you just have to get the reference to the form created in modelMaster/create and use it to renderPartial all */_formInputs you require. Please also remember to include the models for the inputs into the renderPartial-call. So modelMaster/create would look something like:
<?php $form = $this->beginWidget('CActiveForm', array(
'id'=>'foo',
)); ?>
/* Master Inputs here */
// Rendering other models' inputs
<?php $this->renderPartial('model1/_formInputs', array('form' => $form, 'model' => $model1); ?>
<?php $this->renderPartial('model2/_formInputs', array('form' => $form, 'model' => $model2); ?>
/* Render Form Buttons here */
<?php $this->endWidget(); ?>
Submit with Ajax, in Yii it is easy to do and it will keep things easy to understand in the controllers, each controller will have a save and respond with json to confirm the save. There is already ajax validation.
/**
* Performs the AJAX validation.
* #param CModel the model to be validated
*/
protected function performAjaxValidation($model)
{
if(isset($_POST['ajax']) && $_POST['ajax']==='employee-form')
{
$valid = CActiveForm::validate($model);
if(strlen($valid) > 2) {
echo $valid;
Yii::app()->end();
}
}
}
As you can see I have modified it to return the error if there is one (validate returns [] if it is valid, I should probably check for that instead of strlen >2 ), otherwise let the script continue, in this case it will go to the save function.
In Cakephp I have a model called Category and I have another model called Page. Now I connected the Page with $belongsTo to the Category model.
Now I have a form where I can create a new Page:
<?php echo $this->Form->create('Page', array('action' => 'create')); ?>
<?php echo $this->Form->input('title'); ?>
<?php echo $this->Form->input('text'); ?>
<?php echo $this->Form->end('Create new Page'); ?>
Now I want to add the possibility to select the category in the form. I think the solution is simple but I didn't found anything helpful so far...
in your form add this code
echo $this->Form->input('category_id');
now go to your Page controller, inside the appropriate action method, you add this code
$categories = $this->Page->Category->find('list');
$this->set(compact('categories'));
How do I directly input validation as required into input box on ctp file without going to controller or model in Cakephp?
I would use JQuery, with the Jquery Validaton plugin.
It is rather straight foreward to include and write the scripts for.
In the view (using a Users form view for the example)
<?php
echo $form->create('User', array('id'=>'UserForm'));
echo $form->input('User.name', array('class'=>'required', 'minlength'=>2));
echo $form->input('User.email', array('class'=>'required email'));
echo $form->end('Send');
echo $javascript->codeBlock('$("#UserForm").validate();', array('inline'=>true));
In the layout under the header part
echo $javascript->link('jquery-1.6.2.min.js');
echo $javascript->link('jquery.validate.min.js');
In the user controller
var $helpers = array('Html', 'Form', 'Javascript');
I did testrun the code to confirm it works.