I am facing some problems with CakePHP 2.4 in the moment while working with AJAX and JSON.
I want to render data with a view, but save the resulting html as a string in a variable. After that, I would like to set an array, containing this html string among other data to return as JSON object. Unfortunately I didn't find the right way yet.
My controller code so far makes use of the CakePHP json magic:
//Controller (just parts)
$data = $this->paginate();
if($this->request->is('ajax')) {
$jsonResponse = array(
'jobs' => $data,
'foci' => $foci,
'jobTypes' => $jobTypes,
'count_number'=> $count_number
);
$this->set('jsonResponse', $jsonResponse);
$this->set('_serialize', 'jsonResponse');
} else {
// render regular view
$this->set(compact('data', 'foci', 'jobTypes', 'count_number'));
}
This outputs the perfect json in the javascript console, besides the fact, that the data in $data is plain data.
Is it somehow possible, to pass $data to a view, render it, save the output to a string variable $html, and pass $html to jobs in jsonResponse instead of $data?
Yes! You Can render a view into a variable. You just have to create a view object . Inside Your Controller Try This:
$view = new View($this,false);
$view->viewPath='Elements'; // Directory inside view directory to search for .ctp files
$view->layout=false; // if you want to disable layout
$view->set ('variable_name','variable_value'); // set your variables for view here
$html=$view->render('view_name');
// then use this $html for json response
For those of you using CakePhp3
$view = new View($this->request,$this->response,null);
$view->viewPath='MyPath'; // Directory inside view directory to search for .ctp files
$view->layout='ajax'; // layout to use or false to disable
$html=$view->render('view_name');
Don't forget to add this in your namespace
use Cake\View\View;
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