how to return both controller variables and view html contents as ajax json response

Question

I am using CakePHP and using MVC standards.

I am sending a request using ajax and need a response in the same manner as tweeter response, see tweeter sample response below.

Like tweeter sends:

{
"user_id":"155dsfds95",
"html":" \u003Cdiv class=\"module profile-card component profile-header\" data-component-term=\"profile_follow_card\" \u003E\n ",
"screen_name":"abc"
}

I need to send user_id and screen_name from controller and html content from view.

my response data is as follow:

$response['user_id'] = $id;
$response['screen_name'] = 'some_name';

As of now I can send user_id and screen_name as json_encode($response)

But how do I send view content also in data index with the response?

UPDATES

I do not want to make html content in controller, rather it should come from view.

Are you asking how you get a rendered view within a controller? — jeremyharris
– jeremyharris, Commented Jan 8, 2013 at 14:45
Hello jeremyharris, I do not want to make html content in controller due to MVC contraints. How to achieve this like tweeter. — Vijay Choudhary
– Vijay Choudhary, Commented Jan 10, 2013 at 4:59
Have you looked at JSON Views? book.cakephp.org/2.0/en/views/json-and-xml-views.html - assuming you are in CakePHP 2 — David Yell
– David Yell, Commented Jan 23, 2013 at 10:43
Yes david, i have looked at JSON Views, but in json/XML views also, we can only manipulate data coming from controller. What i need is send view content in one json key along with another json key created at controller. — Vijay Choudhary
– Vijay Choudhary, Commented Jan 25, 2013 at 7:01

nanoman · Accepted Answer · 2013-03-11 14:45:22Z

1

You can easily render the current view from your controller and return a JSON response with it.

Working example:

public function index() {
    $response = $this->render();
    $data = array(
        'json_key_1' => 'value1',
        'json_key_2' => 123,
        'action_html' => $response->body()
    );
    $options = array(
        'type' => 'json',
        'body' => json_encode($data)
    );
    return new CakeResponse($options);
}

answered Mar 11, 2013 at 14:45

nanoman

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1 Comment

Vijay Choudhary Over a year ago

Sorry for responding so late, this answer didn't come to my notice and today i accidently seen the answer, when i once again needed for this kind of functionality. +1 and accepting this answer. thanks

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