I'm trying to display an image but is dependent on a dropdown list in Yii. I can get the image from the database and display it, but how to do it dynamically depending on the choice from the dropdown?
Here is the reference: http://www.yiiframework.com/wiki/24/creating-a-dependent-dropdown#hh0 but, let me show you how to do it.
First all all, we need a div where the image will be displayed; I'll create one whose id will be 'img'. Then, the ajax request is specified inside the dropdownlist() as follows:
<?php echo $form->labelEx($model,'attribue'); ?>
<?php echo $form->dropDownList($model,'attribute',
array(/*The options in the DropDownList*/),
array(
'ajax'=>array(
'type'=>'POST',
'url'=>CController::createUrl('YourController/actionWhichEchoesTheImage'),
'update'=>'#img',
)));
?>
<div id="img"> // <---- the result of the ajax call will be displayed here
</div>
In the 'url' attribute we specify the function which will be called when the ajax request triggers. In the 'update' attribute we specified the div where will be displayed the result of calling that function (the image).
Finally, we have to declare the action actionWhichEchoesTheImage(). Let's declare it in the current controller. It would look something like this:
public function actionWhichEchoesTheImage()
{
if(isset($_POST['ModelName']['attribute']))
/*Here goes your code to load the image*/
echo CHtml::image(//Check the reference to see how to set this function);
}
Check CHtml::image() here: http://www.yiiframework.com/doc/api/1.1/CHtml/#image-detail
Related
I've developed a Joomla 2.5 component that does ALMOST exactly what I want it to do. I have a form that contains a fieldset with 2 radio buttons (for "am" and "pm"). I have been unable to figure out how to set the appropriate button's 'checked' attribute based upon other information. It is trivial to set a default within the xml file that defines the form fields, but I don't see how I to do this dynamically.
It is possible? Have I missed something in the documentation that would explain how to do this??
In case you load the form from the view: in the display() method of view.html you will be loading your form:
$this->form = $this->get('Form');
This is invoking a model which in turns extends joomla.application.component.modelform; its getForm method loads the form:
$form = $this->loadForm('com_yourcomp.model', ...
This is what I gather from your description. If this is not the case you might want to move the suggested code below right after you load the form: this is the complete snippet that allows you to set the value of a field:
/// Load the form from the model:
$this->form = $this->get('Form');
// Check for errors.
if (count($errors = $this->get('Errors'))) {
throw new Exception(implode("\n", $errors));
}
//... some logic ...
$this->form->setValue('businessid',null,$businessId);
$this->form->setFieldAttribute( 'businessid', 'readonly', 'true' );
I am using CodeIgniter and I have a controller called contact which passes data to it's view which I have loaded in via the header so that it's on every page, however the data I pass to this view doesn't appear. It only appears if I go straight to the view via the url and I can only presume that this is caused because it is pulled in via another view which has a different controller? Is that right, if so how do I fix it?
For example:
<body>
<div id="header">
<h1>Hello there!</h1>
<?php echo $this->load->view('contact'); ?>
</div>
Are you parsing any data to the 'contact' view? If so, how?
CodeIgniter Userguide - Loading a View
function contact()
{
$data['someinfo'] = "Some Info";
$this->load->view('contact', $data);
}
the problem is calling the 'contact' view from another view doesn't mean the 'contact' controller is being called... that is why you are unable to access the data passed from the 'contact' controller!
To call controller from views you will need https://bitbucket.org/wiredesignz/codeigniter-modular-extensions-hmvc
lets say that i hv this view (main)
<body>
lorem epsim
<div table></div>
lorem epsim
</body>
in controller control1.php i do
$this->load->view('header');
$this->load->view('main',$data);
$this->load->view('footer');
Now i need to load content of div=table from another view (tbl.php),which is called from another controller
control2.php
function load_table(){
$data['x']=1;
$this->load->view('tbl.php',$data);
}
tbl.php view
<ul>$x</ul>
how can i do that ?
i tried to load controler 2 from controller 1 and assign the function load_table to variable and pass that to main view, but it didnt work cuz load->view is executed instead of saving output to variable..
Reason:
i need to do this is that tbl.php view is a complex table that i need to refresh and load via ajax calls, so i need it to be on different view alone
so can some one explain to me how can i work this out ?
You can't call one controller method from another, separate controller. You can, however, get the output of the table view and use that:
// main.php
<body>
lorem epsim
<div table><?php echo $table_content; ?></div>
lorem epsim
</body>
.
// control1.php
$table_data['x'] = 1;
$data['table_content'] = $this->load->view('tbl.php', $table_data, TRUE);
$this->load->view('header');
$this->load->view('main',$data);
$this->load->view('footer');
So, you get the data to pass to the tbl.php view and pass that to the load->view method - also passing TRUE as the third parameter - which will return the contents of that view as a string (instead of outputting that to the browser). Now, you have a $data variable to pass to the main view with the table html included and you can just echo that out in the main view.
How you get the $data['table_content'] data from the view is up to you. You can create another controller method inside control1.php, you can create a helper file that can load the view into a string and return that, etc.
Maybe you can create a view with the table code within and then you can do inside your div for ajax
<div id="for_ajax">
<?php $this->load->view('table'); ?>
</div>
I've similar needs but mine its like a comments wall for issues.
Inside a view use following in a php block
$CI = &get_instance();
$CI->load->view('view_name');
I am new to Yii framework.
I have a form with three fields. I need one of those be a select drop down element that its data comes from previously added data which are in mysql table.
How can I do it ?
If you have a model set up for the table that contains the data you want to use in your dropdown list you can use the CHtml::dropDownList() method the render a dropdown list, and CHtml::listData() to render that model into items for the list, for example;
echo CHtml::dropDownList(
'attribute_name',
'',
CHtml::listData(MyOtherModel::model()->findAll(),'id','name')
);
I use Gii a lot, which uses CActiveForm widget to display forms, if your form uses CActiveForm too you could render your dropdown something like;
$form=$this->beginWidget('CActiveForm', array(
'action'=>Yii::app()->createUrl($this->route),
'method'=>'get',
));
...
echo $form->label($model,'attribute_name');
echo $form->dropDownList(
$model,
'attribute_name',
CHtml::listData(MyOtherModel::model()->findAll(),'id','name')
);
...
$this->endWidget();
Note that CActiveForm uses CHtml::activeDropDownList() rather than CHtml::dropDownList() that I used in my first example, hence the slight difference in syntax between my two examples.
I have an index view which has some elements on it .
index controller code;
$userID = $this->Authsome->get('id');
$qnotes = $this->Qnote->getnotes($userID);
$this->set('qnotes', $qnotes)
$this->render();
elements have been added to the page using
index view code
<?php echo $this->element('lsidebar'); ?>
now the Issue is I also Have an add controller.
add controller code
function add() {
if(!empty($this->data)) {
unset($this->Qnote->Step->validate['qnote_id']);
$this->Qnote->saveAll($this->data);
$this->Session->setFlash('New Note Template has been added.','flash_normal');
}
}
now what I am trying to achieve is once I add a Qnote i want the element('lsidebar') updated
for the new Qnote.
I am Using the Ajax helper. found at http://www.cakephp.bee.pl/
also Here the add qnote View Code :
<?php echo $ajax->submit(
'Submit', array(
'url' => array(
'controller'=>'qnotes',
'action'=>'add')
));
I know its sound like a noob question . can Somebody point me in the right direction atleast.
I have tried everything i could think off. I bet the solution something easy which i didnt think off
help :)
If you want to dynamically update a sidebar with information that is submitted via ajax, there should be a "success" option in your ajax post that would allow you to fire a specific javascript action when the post is finished (or succeeds). You should write a small javascript ajax function to reload the contents of your sidebar when the post succeeds.
See this other stackoverflow answer: CakePHP ajax form submit before and complete will not work for displaying animated gif