Yii CGridView change the header's tr class - yii

So i am working in an app that uses de CGridView extensively but I am porting a webapp from a proprietary framework to Yii. So the CSS files are already written and have been working up until now.
The thing is that in my CGridView widget the headers of columns are enclosed in a TR tag and I have got no clue on where I can add a class attribute to this tag. I've read the documentation and now how to change each header cell individually but not the whole TR.
Thanks for your help!

Ran into a similar problem however as our CSS is for a legacy system I didn't want to roll in yet more CSS rules. In addition I needed support for extra things such as targeting the header with a specific CSS class on the table row and putting in first/last css classes on the items.
To achieve a first/last css on the items you do not need to extend the GridView and can use the handy rowCssClassExpression parameter.
To achieve my second objective of injecting a CSS class into the 'table thead tr' element I did have to override the renderTableHeader() method.
I strongly advise you only consider this route as a last resort because if you update the version of Yii it is conceivable that they make changes that are not backwards compatible with the renderTableHeader() method. Alternatively you could write a test case that runs your widget through a DOM checker to confirm that you only have 1 table element, 1 thead element, 1 tbody element etc...
Yii::import('zii.widgets.grid.CGridView');
class FXSGridView extends CGridView {
public $headerCssClass = 'columnHeadings';
public $itemsCssClass = 'grey';
public $rowCssClassExpression = '$this->rowCssClassFunction($row, $data);';
public $rowCssClass = array('odd','even');
public function rowCssClassFunction($row, $data) {
$classes = array();
if ($row == 0) $classes []= 'first';
if ($row == $this->dataProvider->getItemCount() - 1) $classes []= 'last';
// Do flip/flop on defined rowCssClass
if(is_array($this->rowCssClass) && !empty($this->rowCssClass)) $classes []= $this->rowCssClass[$row % count($this->rowCssClass)];
return empty($classes) ? false : implode(' ', $classes);
}
public function renderTableHeader()
{
if(!$this->hideHeader)
{
echo "<thead>\n";
if($this->filterPosition===self::FILTER_POS_HEADER)
$this->renderFilter();
echo '<tr class="' . $this->headerCssClass . ' ">' . "\n";
foreach($this->columns as $column)
$column->renderHeaderCell();
echo "</tr>\n";
if($this->filterPosition===self::FILTER_POS_BODY)
$this->renderFilter();
echo "</thead>\n";
}
else if($this->filter!==null && ($this->filterPosition===self::FILTER_POS_HEADER || $this->filterPosition===self::FILTER_POS_BODY))
{
echo "<thead>\n";
$this->renderFilter();
echo "</thead>\n";
}
}
}

'columns'=>array(
array(
'name'=>'id',
'header'=>'#',
'htmlOptions'=>array('style'=>'width: 50px; text-align: center;', 'class'=>'zzz'),
'headerHtmlOptions'=>array('class'=>'mytheadclass'),
),
The "headerHtmlOptions" is the one that gives a class to the thead cell of this column.

Unfortunately you cannot do this directly, as there is no provision for adding attributes to the header row tags (see source code).
A straightforward solution would be to subclass CGridView as e.g. MyGridView and override the renderTableHeader method to do what you need it to (add some class variables to MyGridView to let it be configurable). I have used this approach many times in similar situations.

If it's just simple changes you need to make, you might be able to use the generated CSS with something like:
table.admins th {border-right:none;text-align:center;}
table.admins th:first-child {text-align:left;}
etc...
or you could use the Yii-generated ids (view the generated HTML):
<th id="admins-grid_c1">
which may or may not be appropriate depending on how many grids you want to apply the css to and your own naming conventions. You could also use JavaScript/jQuery to manipulate the styles but certainly #Jon's suggestion of creating a custom renderTableHeader is going to give you the most control. (I've also used that approach with lists and renderSorter.)

Related

Remove red * in required fields Yiibooster

I'm using yiibooster and it is really good extension for the frontend, my issues now is that I want to remove the red * that is rendered in the required fields but maintaining the required validator in the model, anyone knows how to do this????
thankss
This is an example of a label generated by a required field validator:
<label for="User_email" class="required">
Email Address <span class="required">*</span>
</label>
Therefore, you can hide it by adding this class to your site's CSS:
span.required {
display: none;
}
If you want to achieve what you want easily, I suggest you to do like below, which is simplest way(in my view point):
Just try to find * selector(the ID or CLASS) name.(using a firebug or any inspector)
Then just do like below in your document.ready():
$(SELECTOR).remove();
NOTES
THE * MIGHT BE CREATED DYNAMICALLY
THIS IS JUST AN SUGGESTION, YOU CAN FIND ANY OTHER POSSIBLE WAYS SUCH AS CHANGING THE CSS CLASS IN ORDER TO DO DISPLAY:NONE OR SOURCE MODIFICATION
<?php echo $form->textFieldGroup($model, 'username',array('label'=>Yii::t('model','Username'))); ?>
or edit line 1223 of TbActiveForm.php from
echo $this->labelEx($model, $attribute, $options['labelOptions']);
to
echo $this->label($model, $attribute, $options['labelOptions']);
Red * is adding according to your validators definition in your model. you have two options.
First in your model add On => 'scenario name' for required validator for the property you want. so you can control the behavior of yii-booster components because they only apply those rules which matches the scenario of the model. for example:
array('password_repeat', 'required', 'on'=>'register'),
It will show Red * only in register scenario (if you set it via $model->setScenario('register');) and in normal times no red * will shown.
Another option for you is when you are creating the form element based on the property marked required by validator rules in model, you can prevent that * from showing but this way will not ignore that validation rule and if you try to submit the form while this form field is empty you will get error from yii (because you just solve showing but in background you have your required validator). for this method, you only need to provide label in your yii-booster form element:
<?php echo $form->textFieldGroup($model,'textField',
array(
'wrapperHtmlOptions' => array(
'class' => 'col-sm-5',
),
'hint' => 'In addition to freeform text, any HTML5 text-based input appears like so.',
>>>>> 'label' => 'Your new value for label which will have no red *',
)
); ?>

How to implement a dynamic js script in Yii List View?

Hello and thanks for reading my question. I have a typical list view:
<?php $this->widget('bootstrap.widgets.TbListView',array(
'dataProvider'=>$dataProvider,
'itemView'=>'_view',
'emptyText'=>'No Jobs',
)); ?>
In my _view file I have a div and a button that slideToggles the div. If I just put the Javascript at the top of the page, it does not work because the results are dynamic and the name of the div changes with the id returned, eg:
id="detailsDiv-<?php echo $data->id_employer_contract;?>"
The problem is in my Javascript, which is as follows:
<?php Yii::app()->clientScript->registerScript('details', "$('#details-$data-id_employer_contract').click(function(){
$('#detailsDiv-$data->id_employer_contract').slideToggle();
return false;});");?>
How can I make this Javascript code dynamic? Meaning, how can I loop through the id? I tried adding the code to the listview property ajaxUpdate but it's still not working. Can someone tell me how I can loop a Javascript in a list view?
Add the id to your toggle buttons as data attribute:
<button class="toggleDetails" data-id="<?php echo $data->id_employer_contract ?>">
Then you can access these data attributes like this js:
<?php Yii::app()->clientScript->registerScript('toggleDetails', "
$('.toggleDetails').click(function(e){
var id = $(this).data('id');
$('#detailsDiv-' + id).slideToggle();
e.preventDefault();
});
", CClientScript::POS_READY) ?>
NOTE: You should not put this javascript into _view.php but into the main file where you render the List View. You only need this one single snippet to deal with all your buttons.

Modifying Yii Portlet Layout

Is it possible to modify the layout of a Yii Portlet?
Within my application I want to use a Portlet to allow for users to admin certain controllers and the Portlet with htmlOptions seems like a perfect fit but I'd like to change the layout/view it uses.
Any advice?
HtmlOptions let you add some options to the container tag.
This widget is not using view.
If you want to modify the presentation of the title, then you should override the method renderDecoration(). Its default code is:
protected function renderDecoration()
{
if($this->title!==null)
{
echo "<div class=\"{$this->decorationCssClass}\">\n";
echo "<div class=\"{$this->titleCssClass}\">{$this->title}</div>\n";
echo "</div>\n";
}
}
To customize the style you can change contentCssClass, decorationCssClass and titleCssClass
Then to change the content of the Portlet you have to override the method renderContent() that is empty by default. You can also set the content as follow:
<?php $this->beginWidget('zii.widgets.CPortlet'); ?>
...insert content here...
<?php $this->endWidget(); ?>

How to customize links for breadcrumbs widget for Yii?

How to customize links for breadcrumbs widget for Yii?
I can change the property tagName, but it for container. I want to change a tag for each links from a to li. How can I do it?
Well there's no setting/property value that you can specify for the default CBreadcrumbs widget class that will give you <li>. You can confirm this if you check out its run() method. In the source you can see this:
public function run(){
// code
foreach($this->links as $label=>$url)
{
if(is_string($label) || is_array($url))
$links[]=CHtml::link($this->encodeLabel ? CHtml::encode($label) : $label, $url);
else
$links[]='<span>'.($this->encodeLabel ? CHtml::encode($url) : $url).'</span>';
}
// more code
}
So what you'll have to do is extend this class and specify your own run method, where you can use <li> or CHtml::tag('li',//other options) instead of the default CHtml::link.
Edit:
Although i would suggest enclosing the <a> for each link with a <li> (instead of replacing it):
$links[]='<li>'.CHtml::link($this->encodeLabel ? CHtml::encode($label) : $label, $url).'</li>';
You can do that with 'activeLinkTemplate' and 'inactiveLinkTemplate' properties:
$this->widget('zii.widgets.CBreadcrumbs',
array('tagName'=>'ul',
'homeLink'=>'<li>'.CHtml::link('Home', array('/')).'</li>',
'activeLinkTemplate'=>'<li>{label}</li>',
'inactiveLinkTemplate'=>'<li class="active">{label}</li>',
'htmlOptions'=> array('class'=>'breadcrumb'),
'separator'=>' / ',
'links'=>$this->breadcrumbs,
));
Or U can write 'homeLink'=>false. But in this case U'll have to set Home url every time with $this->breadcrumbs.

Saving a checkbox value in Yii

I can't figure out how to properly save checkbox values in Yii. I have a MySQL column, active, defined as a tinyint. I have the following form creation code, which correctly shows the checkbox as checked if the value is 1 and unchecked if 0:
<?php echo $form->labelEx($model,'active'); ?>
<?php echo $form->checkBox($model,'active'); ?>
<?php echo $form->error($model,'active'); ?>
And the code to save the form correctly changes other, text-based values:
public function actionUpdate($id)
{
$model=$this->loadModel($id);
if(isset($_POST['Thing']))
{
$model->attributes=$_POST['Thing'];
if($model->save())
$this->redirect(array('thing/index'));
}
$this->render('update',array(
'model'=>$model,
));
}
The value of active is not saved. Where am I going wrong?
You can use htmlOptions array to specify value attribute. Below is the code example:
<?php echo $form->labelEx($model,'active'); ?>
<?php echo $form->checkBox($model,'active', array('value'=>1, 'uncheckValue'=>0)); ?>
<?php echo $form->error($model,'active'); ?>
Since version 1.0.2, a special option named 'uncheckValue' is
available that can be used to specify the value returned when the
checkbox is not checked. By default, this value is '0'.
(This text is taken from YII Documenration)
For every input that you are accepting from user, you need to define it in model::rule(). is active defined there in rule()?
In general, if you are having problems saving to the database, i would replace
$model->save();
with
if($model->save() == false) var_dump($model->errors);
that way, you can see exactly why it did not save. it is usually a validation error.
Please follow:
1. in protected/models/Thing.php add active as a numeric
public function rules()
{
// NOTE: you should only define rules for those attributes that
// will receive user inputs.
return array(
array('active', 'numerical', 'integerOnly'=>true),
//OR optional
array('active', 'safe'),
);
}
Controller action: Its ok
View:
<?php echo $form->labelEx($model,'active'); ?>
<?php echo $form->checkBox($model,'active', array('value'=>1, 'uncheckValue'=>0)); ?>
<?php echo $form->error($model,'active'); ?>
Hope this will work for you...
Article which can be helpful when figuring out how to handle booleans & checkboxes in Yii
http://www.larryullman.com/2010/07/25/handling-checkboxes-in-yii-with-non-boolean-values/
I used a bit type field in my DB and it didn't work.
1.- I changed the field type to tinyint
2.- In the rules function added:
array('active','numerical'),
3.-In the form (as D3K said) do:
<?echo $form->checkBox($model,'active',array('value'=>1, 'uncheckValue'=>0));?>
You can check by printing all the attributes which are being captured. If active is not captured, it must not be safe. you need to declare the variable as safe or define a rule around that variable. This will make the variable safe.
I have similar the same problemce before,I change data type is int,so it save
We can also add a rule as safe in model to pass the values from form to controller without missing.
array('active', 'safe'),
well this post is so old but I've found a solution very useful specially for giving checkbox a value specified rather than number. The new syntax is something like this
notice I'm using ActiveForm
field($model3, 'External_Catering')->checkbox(['id' => 'remember-me-ver', 'custom' => true,'value'=>"External_Catering", 'uncheckValue'=>"vide"]) ?>
1) where my model is =>model3
2) with the name External_Catering
3) that take the value External_Catering and empty when it's uncheckValue
4) in Controller you get the value just by specifying the model and it's attribute like
  $External_Catering=$model3->External_Catering.