in my project i am using the cgridview inside a modal window. now i am facing the problem that when i select the pagination the page get reloaded. how can i load the grid view with ajax pagination. my ui code id given below...
<div id="pick-category-modal" title="Pick Category" class="popupinfoholder2">
<div style="width:700px; height:auto;">
<?php
$widget = $this->widget('zii.widgets.grid.CGridView', array(
'id' => 'response-grid',
'dataProvider' => $pickdataset->pickSection(),
'cssFile' => Yii::app()->baseUrl . '/media/css/gridview.css',
'summaryText' => '',
'ajaxUpdate'=>true,
'enablePagination' => true,
'template' => '{items}',
'pager' => array(
'class' => 'LinkPager',
'cssFile' => false,
'header' => false,
'firstPageLabel' => 'First',
'prevPageLabel' => 'Previous',
'nextPageLabel' => 'Next',
'lastPageLabel' => 'Last',
),
'columns' => array(
.....................
.....................
),));
?>
<div class="grid-view-footer">
<div class="paginationholder">
<div id="pagination">
<?php $widget->renderPager(); ?>
</div>
</div>
</div>
</div>
</div>
Just testing this myself a moment ago, it appears that the pager kicked out with renderPager() does not work with the AJAX setting. The CGridView JavaScript only binds the AJAX to pagers inside the <div> rendered by the widget (#response-grid here), so it does not find and use your additional pager.
I just use the default pager that the CGridView renders at the bottom, and that works fine.
You could override the CGridView JS to fix this, or list it as an enhancement on the Yii bug tracker: http://code.google.com/p/yii/issues/list
Related
I want to apply validation rules only for the input fields that are not empty (not required) for the URL
For example,
If I submit a form and the input is empty, I got a validation error "Instagram Link must be valid URL.", However, I want it without requiring,
and if the input is not empty, I want to apply the rule "valid_url"
How can we fix it?
if (!$this->validate([
'instagram' => [
'rules' => 'valid_url',
'errors' => [
'valid_url' => 'Instagram Link must be valid url.',
],
],
])){
return redirect()->back()->withInput()->with('errors', $this->validator->getErrors());
}
I tried with a permit_link rule, but if I submit it (with an input value like 'mylink' (which is not a valid_url)), it will accept it, but it should not.
Please check the following images and the code:
A form HTML
Result after clicking on edit button
<?= form_open('/settings/edit/1', ['id' => 'setting-form']); ?>
<div class="row">
<div class="col-md-6">
<div class="form-group">
<?= form_label('Instagram'); ?>
<?= form_input(['name' => 'instagram', 'class' => 'form-control', 'id' => 'instagram', 'placeholder' => 'Enter instagram link', 'value' => old('instagram', $setting->instagram)]); ?>
</div>
</div>
</div>
<button type="submit" class="btn btn-dark">Edit</button>
<?= form_close(); ?>
public function edit($id)
{
if (!$this->validate([
'instagram' => [
'rules' => 'permit_empty|valid_url',
'errors' => [
'valid_url' => 'Instagram Link must be valid url.',
],
],
])) {
return redirect()->back()->withInput()->with('errors', $this->validator->getErrors());
}
die('submitted');
}
It should display "Instagram Link must be valid url" and not "submitted",
first
Use separate variable $rules like this :
$rules = ['data' => 'require']
Then
check if $this->request->getVar("instagram"); is empty / is true or etc.. then set it on the $rules
finally
Do Something like this :
$rules = ['extra_data' => 'require'];
if(!empty($this->request->getVar("instagram"))
$rules["instagram"] = "valid_url";
if ($this->validate(rules){
//do something ...
}else {
//do something ...
}
I just noticed that the following examples:
"mylink", "mylink.com", "https://www.mylink.com"
will consider correct for the rule valid_url in Codeigniter (No errors),
While: "https:://www.mylink.com", "mylink#" will apply the validation and the error is applied.
I want a way to create popup form when I press the button using Ajax and bootstrap
and how to create modal for it with bootstrap
use yii\bootstrap\Modal;
you can do it like it, call a view/action in modal
<?php
Modal::begin([
'toggleButton' => [
'label' => '<i class="fa fa-plus"></i> Add',
'class' => 'btn btn-success'
],
'closeButton' => [
'label' => 'Close',
'class' => 'btn btn-danger btn-sm pull-right',
],
'size' => 'modal-lg',
]);
$myModel = new \frontend\models\SomeModel;
echo $this->render('/someview/create', ['model' => $myModel]);
Modal::end();
?>
reference Render Form in popup via AJAX
<div class="form-group">
<?php echo form_textarea(array('name' => 'txtdescription', 'id' => 'desc', 'class' => "ckeditor", 'value' => set_value('txtdescription', htmlspecialchars_decode($desc)))); ?>
</div>
This is my view code when we get data from db
Add 3 argument for set_value.
Try this:
<div class="form-group">
<?php echo form_textarea(array('name' => 'txtdescription', 'id' => 'desc', 'class' => "ckeditor", 'value' => set_value('txtdescription', $desc,false))); ?>
</div>
I want to use PJax in Yii Grid View, not with the associative filter that comes inside the Grid View, but with the Search Filter that's outside it, so it can filter the results inside.
Here is the source of the index file:
<div class="cars-index">
<h1><?= Html::encode($this->title) ?></h1>
<?php echo $this->render('_search', ['model' => $searchModel]); ?>
<p>
<?= Html::a('Create Cars', ['create'], ['class' => 'btn btn-success']) ?>
</p>
<?= GridView::widget([
'dataProvider' => $dataProvider,
'columns' => [
'id',
'name',
['attribute' => 'code',
'label' => 'Colour',
'format' => 'raw',
'value' => 'colour',
'contentOptions' => function ($model, $key, $index, $column){
return ['style' => ' text-align: center; width: 100px;color:white;background-color:#'. $model -> code];
},
'headerOptions' => ['style' => 'text-align: center;'],
],
'price',
['class' => 'yii\grid\ActionColumn'],
],
]); ?>
Am I supposed to create an active form just for the part I want to filter? Or is there another way?
If You can't simply add the filter to you table like this
<?= GridView::widget([
'dataProvider' => $dataProvider,
'filterModel' => $searchModel,
I suggest you use a proper action and a proper search function called by submit based on a specified active form
for action form eg:
<?php $form = ActiveForm::begin([
'id' => 'search-form',
'method' => 'post',
'action' => ['controller/yourAction']
]); ?>
in the controller
$model = new yourActiveForm();
if ($model->load(Yii::$app->request->post()) ) {
$dataProvider = $searchModel->search( [ 'yuorSearchModel'=> ['your_att1' => $model->yourValue1]]);
}
then your render
Conforming to yii2 doc
Pjax only deals with the content enclosed between its begin() and
end() calls, called the body content of the widget. By default, any
link click or form submission (for those forms with data-pjax
attribute) within the body content will trigger an AJAX request. In
responding to the AJAX request, Pjax will send the updated body
content (based on the AJAX request) to the client which will replace
the old content with the new one. The browser's URL will then be
updated using pushState. The whole process requires no reloading of
the layout or resources (js, css).
You may configure $linkSelector to specify which links should trigger
pjax, and configure $formSelector to specify which form submission may
trigger pjax.
You must add the
<?php Pjax::begin(); ?>
.... your active form
<?php Pjax::end(); ?>
and configure the proper $linkSelect and $formSelector
In your filter view:
<div id="myFilter">
<?php $form = ActiveForm::begin([
'id' => 'myFilterForm',
'method' => 'post',
'action' => [...],
]); ?>
...
</div>
And make sure you render filter between Pjax::begin and Pjax::end
Yet here comes the trick. If your server does not respond within default timeout, Pjax gets ignored and page reloaded, so make sure the timeout is big enough:
<?php Pjax::begin([
'id'=>'myGrid',
'timeout' => 10000, // <------------ THIS !!!!!!!
'formSelector' => '#myFilterForm'
]); ?>
<?= $this->render('myFilter', ['model' => $searchModel]); ?>
<?= GridView::widget([
...
]); ?>
<?php Pjax::end(); ?>
Also in your controller, you might want to "reset" the search model, so only data from the request used are attributes actually used by search:
public function actionSearch()
{
$searchModel = new MySearch();
if ($searchModel->load(Yii::$app->request->post())) {
$searchModel = new MySearch(); // "reset"
$dataProvider = $searchModel->search(Yii::$app->request->post());
} else {
$dataProvider = $searchModel->search(Yii::$app->request->queryParams);
}
return $this->render('search', [
'searchModel' => $searchModel,
'dataProvider' => $dataProvider,
]);
}
Hope this helps. Cheers!
I'm just learning Yii a few days ago to developer a website for company. Now i have some issue with Yii pagination and i really need your help.
Currently i can make Yii pagination. However, i need to create a button besides pagination area. Everytime, i click that button, it will call ajax to load next page or previous pages (if current page is last page).
Here is my model:
public function getListDesc()
{
$arrPlace = array(0 => '--- --- --- ---');
$criteria=new CDbCriteria;
$criteria->select='t.*,pi.path path';
$criteria->join='JOIN places_images pi ON t.id=pi.places_id';
$criteria->group = 't.id';
$criteria->compare('t.deleted', '<>1', false);
$criteria->compare('t.status',1);
$criteria->compare('outlet','Outlet 1');
$criteria->order = 't.id DESC';
return new CActiveDataProvider('Places', array(
'criteria'=>$criteria,
'pagination' => array(
'pageSize' => 2
)
));
}
Here is a part of my index page which is call that model:
<?php
$this->widget('zii.widgets.CListView', array(
'id' => 'places-list-right',
'dataProvider'=> Places::model()->getListAsc(),
'summaryText'=>'',
'pager' => array(
'header' => '',
'prevPageLabel' => '<<',
'nextPageLabel' => '>>',
),
'itemView'=>'_fea_oulet',)
);
?>
Here is _fea_oulet.php file:
<div class="place_img" id="<?=$data->id?>">
<div class="plusimage">
<a href="javascript:void(0);" onclick="randnew();">
<img alt="<?=$data->id?>" src="<?php echo Yii::app()->request->baseUrl; ?>/images/9life-show-more-images.png" style="border:none;border-radius:0;-webkit-border-radius: 0; -moz-border-radius:0">
</a>
</div>
<?php echo CHtml::image("timthumb.php?src=".$data->path."&w=294&h=294&zc=1&q=100",$data->title);?>
<div class="featitle">
<h3><span><?php
if(strlen($data->title) > 30) {
echo CHtml::encode(mb_substr($data->title,0,30,'UTF-8')) . '...';
}
else{
echo CHtml::encode($data->title);
}
?></span>
<img src="<?php echo Yii::app()->request->baseUrl; ?>/images/9life-show-more-green.png" style="border:none;border-radius:0;-webkit-border-radius: 0; -moz-border-radius:0">
</h3>
</div>
</div>
Any help are appreciates.
Thanks,
You can override CLinkPager class and use your own implementation MyLinkPager in your CListView widget's call specifying its pager property like this:
$this->widget('zii.widgets.CListView', array(
'id' => 'places-list-right',
'dataProvider' => Places::model()->getListAsc(),
'summaryText' => '',
'pager' => array(
'class' => 'MyLinkPager',
'header' => '',
'prevPageLabel' => '<<',
'nextPageLabel' => '>>',
),
'itemView' => '_fea_oulet'
));
run() method of CLinkPager is the place where pager buttons' rendering happens so you can override this method directly. Or you can override createPageButtons() method by adding your custom button to $buttons array if you want to see it along with other page buttons.