Get {} as an output instead of [] - php

Question

I have service which returns the menu of venues, but if menu is not exist, I should return {} as an output, otherwise ios app cannot parse the response and app destroys.
Now reponse looks like:
{
"response": []
}
I should have
{
"response": {}
}
API services programmed with PHP.
When menu empty url: http://ilovejetset.com/api/v2/menu/index/422
When some menu exist: http://ilovejetset.com/api/v2/menu/index/423
The code for creating response:
in 'after' function:
$this->response->body(json_encode(array('response' => $this->response_json)));
when no menu:
$this->response_json = array();

Answer 1

Check out this page:
Predefined Constants (JSON) More specifically JSON_FORCE_OBJECT
If the data which is deeper than responses needs to be an array then you will need to loop over first encode all of that data, then encode the top layer.
$test = array(
"responses" => array()
);
echo json_encode($test, JSON_FORCE_OBJECT);

Answer 2

If you need an object, then you need to create an object in PHP as well. As long as you're using arrays, PHP will preferably encode them as JSON arrays.
$data = new stdClass;
$data->foo = 'bar';
echo json_encode(array($data));
or:
echo json_encode(array((object)array('foo' => 'bar')));

Answer 3

if you use json_encode this function, you can try to add the option JSON_FORCE_OBJECT as fllows.
<?php
$var = array();
$var['response'] = array();
echo json_encode($var, JSON_FORCE_OBJECT);
$var1 = array();
$var1['response'] = array();
$var1['response']['name'] = 'jediliang';
echo json_encode($var1, JSON_FORCE_OBJECT);
?>
the output looks like this:
{"response":{}}{"response":{"name":"jediliang"}}

Answer 4

For more clarification try this following example
class a {
var $name;
};
$a = new a;
$a->name = array("name"=>"fname");
echo json_encode(array("a"=> $a, "name"=>array("fname","lname")));
Hope it clears all your doubts