So I have this part in my View:
<body>
<div id = "content">
<?php echo $catalog ?>
</div>
</body>
There are also other variables in it. Here is the part of my Controller where I send them to the View:
$this->load->view('layout',array(
'categories' => $categories,
'home_menu' => $home_menu,
'information' => $information,
'favourite' => $favourite,
'new_products' => $new_products,
'bestsellers' => $bestsellers,
'login_info' => $login_info,
'catalog' => ''
));
I want to create second controller, which when activated sends a second view to the variable $catalog.
Something like this (similar to Kohana):
$this->layout->catalog = $this->load->view('products/catalog', array(
'name' => $name,
'description' => $description));
But it's not working.
My question is, how can I show this second nested view after clicking on a link that activates the second Controller?
EDIT:
But I want to send the catalog view to $catalog variable after the user has clicked on a link that activates second controller, which look something like this:
$products = $this->Product_model->list_products($category_id);
foreach ($products as $row)
{
$name = $row->name;
$description = $row->description;
}
.. after that I want $name and $description to be passed to:
$this->load->view('products/catalog', array(
'name' => $name,
'description' => $description));
..which itself to be passed to $catalog in the layout view defined in the first controller
You can call a $this->load->view within the view's code but I would not recommend it.
Instead pass true as the 3rd parameter in the load view function and this will return the view rather than echo it straight out. Then you can assign that returned code to your original view.
I'm hoping I'm understanding your question fully, but if not, I apologize.
My guess is that you're loading the page with all of the 'extras' and want to be able to update the 'content' part of your page through a user initiated click.
If you're implementing a javascript based solution, then you just need a controller that will output the html fragment and inject that into the current page via an ajax call.
If you're not implementing javascript, then it would be an entire page refresh, so you would just rebuild the page and pass the selected catalog content to the controller.
UPDATE
To do this without ajax or hmvc, you need to get the contents from another controller into this controller, so you could just make an additional request with php:
$catalog_content = file_get_contents('/url_to_second_controller.html');
$this->load->view('layout',array(
'categories' => $categories,
'home_menu' => $home_menu,
'information' => $information,
'favourite' => $favourite,
'new_products' => $new_products,
'bestsellers' => $bestsellers,
'login_info' => $login_info,
'catalog' => $catalog_content
));
Related
I have following code:
return array(
ULogt::UPDATE => '
<div>
Navigate
</div>
'
This code locates in the class called links.php.
I need to navigate to ('viewform', array('model'=>$this->loadJson($id)), when the user presses Navigate button. I do not know how to insert this code instead of #link. How can I do it?
CHtml::link(
"Navigate",
"javascript:void(0);", // link for destination or you can handle same with jQuery
array(
'id' => 'navigation-id', // id for handeling in jQuery
'key' => $data, // Required data will be appeared as attributes for this link
)
);
You can create link like
Yii::app()->createUrl(
'/profile/membership/view', // your link
array(
'id'=> 1 // data to be sent
)
)
Check for URL formating
I need to load a view into a view within CodeIgniter, but cant seem to get it to work.
I have a loop. I need to place that loop within multiple views (same data different pages). So I have the loop by itself, as a view, to receive the array from the controller and display the data.
But the issue is the array is not available to the second view, its empty
The second view loads fine, but the array $due_check_data is empty
SO, I've tried many things, but according to the docs I can do something like this:
Controller:
// gather data for view
$view_data = array(
'loop' => $this->load->view('checks/include/due_checks_table', $due_check_data, TRUE),
'check_cats' => $this->check_model->get_check_cats(),
'page_title' => 'Due Checks & Tests'
);
$this->load->view('checks/due_checks',$view_data);
But the array variable $due_check_data is empty
I'm just getting this error, saying the variable is empty?
Message: Undefined variable: due_check_data
You are passing the $view_data array to your view. Then, in your view, you can access only the variables contained in $view_data:
$loop
$check_cats
$page_title
There is no variable due_check_data in the view.
EDIT
The first view is contained in the variable $loop, so you can just print it in the second view (checks/due_checks):
echo $loop;
If you really want to have the $due_check_data array in the second view, why don't you simply pass it?
$view_data = array(
'loop' => $this->load->view('checks/include/due_checks_table', $due_check_data, TRUE),
'check_cats' => $this->check_model->get_check_cats(),
'page_title' => 'Due Checks & Tests',
'due_check_data' => $due_check_data
);
$this->load->view('checks/due_checks',$view_data);
Controller seems has no error. Check out some notices yourself:
<?=$due_check_data?>
This only available in PHP >= 5.4
<? echo $due_check_data; ?>
This only available when you enable short open tag in php.ini file but not recommended
You are missing <?php. Should be something like this
<?php echo $due_check_data; ?>
OK, i managed to solve this by declaring the variables globally, so they are available to all views.
// gather data for view
$view_data = array(
'due_check_data' => $combined_checks,
'check_cats' => $this->check_model->get_check_cats(),
'page_title' => 'Due Checks & Tests'
);
$this->load->vars($view_data);
$this->load->view('checks/due_checks');
Assume I'm in my items controller.
Ok say I am in my view action (the url would be something like /items/view/10012?date=2013-09-30) which lists a list of items that belongs to a client on a given date.
I want to link to add a new item. I would use the htmlhelper like so:
echo $this->Html('action'=>'add');
In my add action I have a form which has fields like client_id and item_date.
When I'm in my view action I know these values as I am viewing the items for a specific client on a specific date. I want to pass these variables to my add action so it will prefill those fields on the form.
If I add a query string in my link ('?' => array('client_id'=>$client_id)) it breaks the add action as it will give an error if the request is not POST. If I use a form->postLink I get another error as the add action's POST data must only be used for adding the record, not passing data to prefill the form.
I basically want to make my link on the view page pass those 2 variables to the add action in the controller so I can define some variables to prefill the form. Is there a way to do this?
Here is my add controller code. It may differ in content a bit from my question above as I have tried to simplify the question a bit but the concept should still apply.
public function add(){
if ($this->request->is('post')) {
$this->Holding->create();
if ($this->Holding->save($this->request->data)) {
$this->Session->setFlash(__('Holding has been saved.'), 'default', array('class' => 'alert alert-success'));
return $this->redirect(array('action' => 'index'));
}
$this->Session->setFlash(__('Unable to add your holding.'), 'default', array('class' => 'alert alert-danger'));
}
$this->set('accounts', $this->Holding->Account->find('list'));
$sedol_list = $this->Holding->Sedol->find('all', array(
'fields' => array(
'id', 'sedol_description'
),
'recursive' => 0,
'order' => 'description'
)
);
$this->set('sedols', Hash::combine($sedol_list, '{n}.Sedol.id', '{n}.Sedol.sedol_description') );
}
Why not use proper Cake URL parameters?
echo $this->Html->link('Add Item', array(
'action' => 'add',
$client_id,
$item_date
));
This will give you a much nicer URL like:
http://www.example.com/items/add/10012/2013-09-30
And then in your controller, you modify the function to receive those parameters:
public function add($client_id, $item_date) {
// Prefill the form on this page by manually setting the values
// in the request data array. This is what Cake uses to populate
// the form inputs on your page.
if (empty($this->request->data)) {
$this->request->data['Item']['client_id'] = $client_id;
$this->request->data['Item']['item_date'] = $item_date;
} else {
// In here process the form data normally from when the
// user has submitted it themselves...
}
}
In both cakephp-1.2 and cakephp-1.3 I have used the following code snippet in an element named head called from the blog layout:
$this->preMetaValues = array(
'title' => __('SiteTitle', true).' '.$title_for_layout,
'desc' => Configure::read('siteTitle').', '.Configure::read('siteSlogan'),
'keywords' => Configure::read('keywords'),
'type' => 'article',
'site_name' => __('SiteTitle', true),
'imageURL' => $html->url('/img/logo.png', true)
);
if(!isset($this->metaValues)){
$this->metaValues = $this->preMetaValues;
}
else{
$this->metaValues = array_merge($this->preMetaValues, $this->metaValues);
}
<?php echo $html->meta('description',$this->metaValues['desc']); ?>
<?php echo $html->meta('keywords', $this->metaValues['keywords']);?>
I used the above code to define or modify meta-tags values from the any view file. The preMetaValues is regarded as the default values. If there is any metaValues defined in the view, this code will modify it and make the metaValues ready to be used.
Now with cakephp-2.4, the described code generates the following error:
Helper class metaValuesHelper could not be found.
Error: An Internal Error Has Occurred.
Indeed, I don't know why CakePHP regards this variable as helper? and how could I fix this issue?
You can do it by setting the variable from your controller action:
$this->set('title_for_layout', 'Your title');
And then in the view, printing it with:
<title><?php echo $title_for_layout?></title>
You have an example of this at the documentation:
http://book.cakephp.org/2.0/en/views.html#layouts
Just treat them as any other variable.
Why you're using $this object? Can't you use a simple solution like this:
$preMetaValues = array(
'title' => __('SiteTitle', true).' '.$title_for_layout,
'desc' => Configure::read('siteTitle').', '.Configure::read('siteSlogan'),
'keywords' => Configure::read('keywords'),
'type' => 'article',
'site_name' => __('SiteTitle', true),
'imageURL' => $html->url('/img/logo.png', true)
);
if(!isset($metaValues)){
$metaValues = $preMetaValues;
}
else{
$metaValues = array_merge($preMetaValues, $metaValues);
}
<?php echo $html->meta('description',$metaValues['desc']); ?>
<?php echo $html->meta('keywords', $metaValues['keywords']);?>
Finally I have found the solution. It is simply about how to set a variable for the layout from a view. It seems that in earlier versions of cakephp the view was processed before the layout while now in cakephp-2.4 the layout is processed first, so any override of any variable defined in the layout from the view will not success.
Hence, the solution will depend on the set method of the view object something as follows:
//in some view such as index.ctp
$this->set('metaValues', array(
'title', 'The title string...',
'desc' => 'The description string...'
)
);
Also as Alvaro regarded in his answer, I have to access those variable without $this, i.e as local variables.
This answer is inspired from: Pass a variable from view to layout in CakePHP - or where else to put this logic?
i've been using Yii framework for some time now, and i've been really having a good time especially with these widgets that makes the development easier. I'm using Yii bootsrap for my extensions..but i'm having a little trouble understanding how each widget works.
My question is how do i display the widget say a TbDetailView inside a tab?
i basically want to display contents in tab forms..however some of them are in table forms...some are in lists, detailviews etc.
I have this widget :
$this->widget('bootstrap.widgets.TbDetailView',array(
'data'=>$model,
'attributes'=>$attributes1,
));
that i want to put inside a tab
$this->widget('bootstrap.widgets.TbWizard', array(
'tabs' => $tabs,
'type' => 'tabs', // 'tabs' or 'pills'
'options' => array(
'onTabShow' => 'js:function(tab, navigation, index) {
var $total = navigation.find("li").length;
var $current = index+1;
var $percent = ($current/$total) * 100;
$("#wizard-bar > .bar").css({width:$percent+"%"});
}',
),
and my $tabs array is declared like this :
$tabs = array('studydetails' =>
array(
'id'=>'f1study-create-studydetails',
'label' => 'Study Details',
'content' =>//what do i put here?),
...
...);
when i store the widget inside a variable like a $table = $this->widget('boots....);
and use the $table variable for the 'content' parameter i get an error message like:
Object of class TbDetailView could not be converted to string
I don't quite seem to understand how this works...i need help..Thanks :)
You can use a renderPartial() directly in your content, like this:
'content'=>$this->renderPartial('_tabpage1', [] ,true),
Now yii will try to render a file called '_tabpage1.php' which should be in the same folder as the view rendering the wizard. You must return what renderPartial generates instead of rendering it directly, thus set the 3rd parameter to true.
The third parameter that the widget() function takes is used to capture output into a variable like you are trying to do.
from the docs:
public mixed widget(string $className, array $properties=array ( ), boolean $captureOutput=false)
$this->widget('class', array(options), true)
Right now you are capturing the object itself in the variable trying to echo out an object. Echo only works for things that can be cast to a string.