radioButtonList checked by default - yii

I created a option list with 2 options: Yes and No like below
<?php echo $form->radioButtonList($model,'active', array(1=>'Yes', 0=>'No'), array('separator'=>"" )); ?>
How can I set option 1 to be selected by default ?

You have to set $model->active = 1 in your controller.

I prefer on view page, just before the form element. as
<?php $model->isNewRecord ? $model->active = 1: $model->active = $model->active ; ?>
This will take care of Update action also.

You can just also set a default value in the Model itself:
Here's a form where I use a radioButtonList for reportType and have one selected by default:
class FreeReportForm extends CFormModel
{
public $userId;
public $email;
public $callId;
public $reportType = 1;
public $companyNumber;
public $expiry;
...
}

you should add a line "$model->active=1" in your controller's function actionXXX().
for example , if the radio is in the create page, there must be a function named actionCreate() and that is the very place you add the code.

You can set value 1 as default selected value without using $model also
<?php echo $form->radioButtonList($model,'1', array(1=>'Yes', 0=>'No'), array('separator'=>"" )); ?>

Related

Shorter way of displaying ActiveRecord object in view than using "attributes" array

My Model returns this to my Controller:
namespace app\models;
use yii\data\ActiveDataProvider;
use \yii\db\ActiveRecord;
class Questions extends ActiveRecord{
public static function getQuestionById($id){
return Questions::find()->select('title, is_textarea')->where(['id'=>$id])->one();
}
public static function model($className=__CLASS__)
{
return parent::instance();
}
}
In the Controller, it's sent to the View:
return $this->render('question', ['question' => Question::getQuestionById($id)]);
Then in my View, I must display the question like such:
<?= $question['attributes']['title'] ?>
Isn't there a way to display in a more human-friendly way, for example:
<?= $question->title ?>
I tried it and it's throwing a "Trying to get property of non-object" exception. I thought that was the way to go in Yii, just like Django, CodeIgniter, etc but I can't find an example in the documentation how to display model data in views
While sending to the view you are assigning ['questions'=>$question] so using
$questions->title
should work
To avoid "Trying to get property of non-object" exception you should check that $question exists
Controller
public function actionTest($id)
{
$question = Question::getQuestionById($id);
if (empty($question)) {
throw new \yii\web\NotFoundHttpException('Question not found');
}
return $this->render('test', [
'question' => $question,
]);
}
View
<?php
use yii\helpers\Html;
use yii\web\View;
use app\models\Question;
/**
* #var $this View
* #var $question Question
*/
?>
<div>
<h1><?= Html::encode($question->title) ?></h1>
<p>
<?= Html::encode($question->is_textarea) ?>
</p>
</div>
I think you should clarify which are the steps you need to achieve what you want:
Pass the data to your view. In your case, you want to pass a model of Questions. You don't need to pass data from the model to the controller. Actually, the data you need is the model itself.
So you can simply do this inside your controller's action:
public function actionTest($id)
{
$question = Questions::findOne(['id' => $id]);
return $this->render('test', [
'question' => $question,
]);
}
Display the data. Now in your view file you simply access to the model fields that you need:
<?= $question->title ?>

Set successCssClass and errorCssClass for all input by default

I want set successCssClass and errorCssClass all form's inputs in Yii. How I can doing it.
Unfortunately the the defaults are defined in jquery active form plugin i.e /framework/web/js/jquery.yiiactiveform.js look at the end .
However,as #Michiel suggested, You can extend CActiveForm sth like this:
MyActiveForm extends CActiveForm{
public function error($model,$attribute,$htmlOptions=array(),$enableAjaxValidation=true,$enableClientValidation=true)
{
//my defaults
$defaults=array('errorCssClass'=>'myErrorCss',
'successCssClass'=>'mySuccessCss',
'validatingCssClass'=>'myValidationCss');
$htmlOptions=array_merge($defaults,$htmlOptions);
return parent::error($model,$htmlOptions,$enableAjaxValidation,$enableClientValidation);
}
}
And you have to use your new class e.g
<?php $form = $this->beginWidget('MyActiveForm', array(
..
?>
I don't see a better way right now. Either you have extend jquery.yiiactiveform plugin or CActiveForm .

Yii 1.1 - creating a multi step form with validation

I'm basically trying to create a multi-step form using the CActiveForm class in Yii. The idea is I want to use the built-in functionality to achieve this in the simplest way possible. The requirement I have is as follows:
A multi step ONE PAGE form (using DIVs that show/hide with jQuery)
AJAX validation on EACH step (validate step-specific attributes only)
The validation MUST work using the validateOnChange() and validateOnSubmit() methods
This is a half-working solution I have developed so far:
View:
<div class="form">
<?php $form = $this->beginWidget('CActiveForm', array(
'id'=>'listing-form',
'enableClientValidation'=>false,
'enableAjaxValidation'=>true,
'clientOptions'=>array(
'validateOnChange'=>true,
'validateOnSubmit'=>true,
'afterValidate'=>'js:validateListing',
),
)); ?>
<?php echo $form->errorSummary($model); ?>
<div class="step" id="step-1">
// model input fields
<?php echo CHtml::submitButton('Next Step', array('name'=>'step1')); ?>
</div>
<div class="step" id="step-2" style="display: none;">
// model input fields
<?php echo CHtml::submitButton('Next Step', array('name'=>'step2')); ?>
</div>
<div class="step" id="step-3" style="display: none;">
// model input fields
<?php echo CHtml::submitButton('Submit', array('name'=>'step3')); ?>
</div>
<?php $this->endWidget(); ?>
</div>
JavaScript:
function validateListing(form, data, hasError)
{
if(hasError)
{
// display JS flash message
}
else
{
if($('#step-1').css('display') != 'none')
{
$('#step-1').hide();
$('#step-2').show();
}
else if($('#step-2').css('display') != 'none')
{
$('#step-2').hide();
$('#step-3').show();
}
else if($('#step-3').css('display') != 'none')
{
return true; // trigger default form submit
}
}
}
Controller:
public function actionCreate()
{
$model = new Listing;
// step 1 ajax validation
if(isset($_POST['step1']))
{
$attributes = array('name', 'address1', 'etc');
$this->performAjaxValidation($model, $attributes);
}
// step 2 ajax validation
if(isset($_POST['step2']))
{
$attributes = array('category', 'type', 'etc');
$this->performAjaxValidation($model, $attributes);
}
// step 3 ajax validation
if(isset($_POST['step3']))
{
$attributes = array('details', 'source', 'etc');
$this->performAjaxValidation($model, $attributes);
}
// process regular POST
if(isset($_POST['Listing']))
{
$model->attributes = $_POST['Listing'];
if($model->validate()) // validate all attributes again to be sure
{
// perform save actions, redirect, etc
}
}
$this->render('create', array(
'model'=>$model,
));
}
protected function performAjaxValidation($model, $attributes=null)
{
if(isset($_POST['ajax']) && $_POST['ajax']==='listing-form')
{
echo CActiveForm::validate($model, $attributes);
Yii::app()->end();
}
}
To summarise. Basically what I have is a form with 3 submit buttons (one for each step). In my controller I check which submit button was pressed and I run AJAX validation for the attributes specific to that step.
I use a custom afterValidate() function to show/hide the steps upon submit. On step 3, the default form submit is triggered, which posts all the form attributes to the controller.
This works well, except it won't work with validateOnChange() (since the submit button doesn't get posted). Also I was wondering whether this is actually the best way to do this, or if anyone knows of a better way?
Thanks.
I'd suggesting using scenarios to turn on and off the appropriate rules. Adjust the model scenario based on what is sent to your controller.
Note: this may also be a really good place to use a CFormModel instead of a CActiveRecord, depending on what is in your form.
Edit: can you add a hidden field to each div section that contains the info about what step you are on? Seems like that should work instead of your submit buttons.
OPTION 1
When you do not receive a button, why not validate the entire form, why do you need to validate only specific attributes? Yii will validate the entire model, send back all the errors but only that particular error will be shown by the active form because that is how it works already.
OPTION 2
You can have 3 forms (not 1 like you have now), 1 on each step. Also create 3 scenarios 1 for each step.
Each form has a hidden field that gets posted with the form, it can actually be the scenario name just validate it when it comes in. Validate the model using this hidden field to set the scenario you are on.
You can cache parts on the model when the form is submitted successfully and at the end you have the complete model.
you can always have custom validation and it won't break your normal form validation
in your model
private $step1 = false;
private $step2 = false;
private $all_ok = false;
protected function beforeValidate()
{
if(!empty($this->attr1) && $this->attr2) // if the fields you are looking for are filled, let it go to next
{
$this->step1 = true;
}
if($this->step1)
{
... some more validation
$this->step2 = true;
}
if($this->step2)
{
... if all your logic meets
$this->all_ok = true;
}
// if all fields that your looking for are filled, let parent validate them all
// if they don't go with their original rules, parent will notify
if($this->all_ok)
return parent::beforeValidate();
$this->addError($this->tableSchema->primaryKey, 'please fillout the form correctly');
return false;
}
I think better create specific class for each step of validation and use scenarios with rules. Below is small example.
//protected/extensions/validators
class StepOneMyModelValidator extends CValidator
{
/**
* #inheritdoc
*/
protected function validateAttribute($object, $attribute)
{
/* #var $object YourModel */
// validation step 1 here.
if (exist_problems) {
$object->addError($attribute, 'step1 is failed');
}
...
Create other classes(steps) for validation...
// in your model
public function rules()
{
return array(
array('attr', 'ext.validators.StepOneMyModelValidator', 'on' => 'step1'),
...
How to use in controller:
$model = new Listing();
$steps = array('step1', 'step2', /* etc... */);
foreach($_POST as $key => $val) {
if (in_array($key, $steps)) {
$model->setScenario($key);
break;
}
}
$model->validate();
echo '<pre>';
print_r($model->getErrors());
echo '</pre>';
die();
Or we can validate all steps in one validator.

CGridview filter on page load with pre define value in search field

I am working with the Yii framework.
I have set a value in one of my cgridview filter fields using:
Here is my jQuery to assign a value to the searchfield:
$('#gridviewid').find('input[type=text],textarea,select').filter(':visible:first').val('".$_GET['value']."');
And here my PHP for calling the cgridview:
$this->widget('zii.widgets.grid.CGridView', array(
'id'=>'bills-grid',
'dataProvider'=>$dataProvider,
'filter'=>$model,
'cssFile'=>Yii::app()->baseUrl . '/css/gridview.css',
'pager'=>array(
'class'=>'AjaxList',
'maxButtonCount'=>25,
'header'=>''
),
'columns' => $dialog->columns(),
'template'=>"<div class=\"tools\">".$dialog->link()." ".CHtml::link($xcel.' Export to excel', array('ExcelAll'))."</div><br />{items}{summary}<div class=\"pager-fix\">{pager}</div>",));
The value appears in the search field and my cgridview works correctly without any issues, but I am unable to trigger the cgridview to refresh or filter. Does anyone know who to trigger the cgridview to filter after page load with a predefined value?
Any help would be greatly appreciated and please let me know if you need additional information.
Thank you.
You can solve the problem without any clientside code modification. In your controller action just set the default value for the attribute as shown below
public function actionAdmin()
{
$model = new Bills();
$model->unsetAttributes();
$model->attribute_name="default filter value";//where attribute_name is the attribute for which you want the default value in the filter search field
if(isset($_GET['Bills'])){
$model->attributes = $_GET['Bills'];
}
$this->render('admin',array('model'=>$model));
}
Have a look at 'default' index action that gii generates:
public function actionIndex()
{
$model = new Bills();
$model->unsetAttributes();
if(isset($_GET['Bills'])){
$model->attributes = $_GET['Bills'];
}
$this->render('index',array('model'=>$model));
}
So if you add one line like: $model->attribute = 'test';, you're done. 'attribute' is of course the attribute that has to have the default filter value (in this case value is 'test') :). So your code looks like:
public function actionIndex()
{
$model = new Bills();
$model->unsetAttributes();
if(isset($_GET['Bills'])){
$model->attributes = $_GET['Bills'];
}
if(!isset($_GET['Bills']['attribute']) {
$model->attribute = 'test';
}
$this->render('index',array('model'=>$model));
}
Of course youre attribute will have a test value (in filter) set up as long as you wont type anything in its filter field. I hope that that's what you're looking for. Your filter should work as always.
Sorry for my bad english :)
Regards
You can use Yii's update:
$.fn.yiiGridView.update('bills-grid', {
type: 'GET',
url: <?php echo Yii::app()->createUrl('controller/action') ?>"?Class[attribute]=<?php echo $_GET['value'] ?>
success: function() {
$.fn.yiiGridView.update('bills-grid');
}
});
This is how i do it, just change the URL, it should be the same controller action of the gridview and change URL parameters to the structure represented in there, should be like Bills[attribute]=value.

Data saved as null when using RadioButtonList or dropDownList

I'm new to yii. I have this problem where my data stored in radiobuttonlist or dropDownList is not saved in the database. It always shows as null. here's my code
View:
<?php
$form = $this->beginWidget('CActiveForm');
echo $form->label($model,'gender');
echo $form->radioButtonList($model,'gender',array('M'=>'Male','F'=>'Female'));
echo $form->label($model,'cat');
echo $form->dropDownList($model,'cat',$category);
echo CHtml::submitButton('Submit');
$this->endWidget();
?>
Controller:
public function actionCreate()
{
$model=new Test;
if(isset($_POST['Test']))
{
$model->attributes=$_POST['Test'];
if($model->save()){
$this->redirect(array('index'));
}
else
var_dump($model->errors);
}
$cat = array('st'=>'STAFF','ot'=>'OTHERS');
$model->gender='M';
$this->render('create',array(
'model'=>$model,'category'=>$cat
));
}
Kindly help... Thanks in advance
EDIT: After adding the required in the rule section it works like a charm
Well here's the modified Test model
public function rules()
{
return array(
array('gender,cat', 'required'),
array('name', 'length', 'max'=>45),
);
}
Post your model here.I think your problem is in Test model.
I see you solved it using 'required', but if there are some fields that are not mandatory, you can just use the 'safe' rule. The point is that every attribute of your form has to be on the rules of your model.
Have a look to Understanding "Safe" Validation Rules.