CakePHP - How to return string (like JSON) from controller action to Ajax request

Question

So I have my JavaScript making an Ajax call to /my_controller/ajax_action but then in the controller I don't know what to do to output something back to the JavaScript.

I am getting errors because there is no view for MyController::ajaxAction() but obviously there is no view for it, so what do I do?

What if the response isn't JSON? I would like a solution that doesn't just return JSON strings, but anything. — BadHorsie
– BadHorsie, Commented Aug 19, 2011 at 16:45
You should always return JSON if you ask for JSON (via .json extension) for example. What else would it be? You can pass pretty much anything with JSON. See this article on details how to achieve that. — mark
– mark, Commented Jul 10, 2014 at 23:18

gyger · Accepted Answer · 2012-02-26 23:00:40Z

8

do this, have your variables you want to output in an array let's say $data, then pass that array to the view using the $this->set('data', $data); method, then create a view /General/SerializeJson.ctp. In that view file, put <?PHP echo json_encode($data); ?> after that you can use $this->render('/General/SerializeJson'); and it should output the json.

General code...

/Controllers/MyController.php

public class MyController extends AppController
{
    public function ajaxAction()
    {
        $data = Array(
            "name" => "Saad Imran",
            "age" => 19
        );
        $this->set('data', $data);
        $this->render('/General/SerializeJson/');
    }
}

/Views/General/SerializeJson.ctp

<?PHP echo json_encode($data); ?>

edited Feb 26, 2012 at 23:00

gyger

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answered Aug 19, 2011 at 16:45

Saad Imran.

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Comments

Shaz Ravenswood · Accepted Answer · 2011-08-21 12:40:42Z

4

Easiest way I found was to disable the automatic rendering:

function ajax_action($data = null) {
    if($this->RequestHandler->isAjax()) {
        $this->autoRender = false;
        //process my data and return it
        return $data;
    } else {    
        $this->Session->setFlash(__('Not an AJAX Query', true));
        $this->redirect(array('action' => 'index'));
    }   
}

answered Aug 21, 2011 at 12:40

Shaz Ravenswood

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Comments

thecodeparadox · Accepted Answer · 2011-08-19 16:43:37Z

3

try this:

within your view folder for the corresponding controller(my_controller) make a folder named json and place a file named index.ctp and within that ctp file write this code:

<?php echo json_encode($yourVariableNameReturnedFromController); ?>

within your my_controller in index() wrote this code:

$this->set('yourVariableNameReturnedFromController', $this->YOURMODEL->find('all'));

within your app_controller.php(if not exist you have to made it) write this code

function beforeFilter(){
  if ($this->RequestHandler->ext == 'json') {
     Configure::write('debug', 0);
  }
}

answered Aug 19, 2011 at 16:43

thecodeparadox

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1 Comment

thecodeparadox Over a year ago

here i use index() function for accessing the json and your url will like '/my_controller/index.json'

Guilherme Oliveira · Accepted Answer · 2014-07-10 20:53:00Z

1

AutoRender=false and Return json_encode($code)

public function returningJsonData($estado_id){
    $this->autoRender = false;

    return json_encode($this->ModelBla->find('first',array(
        'conditions'=>array('Bla.bla_child_id'=>$estado_id)
    )));
}

answered Jul 10, 2014 at 20:53

Guilherme Oliveira

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