How do you get the string value of a MongoID using PHP?

Question

After doing an insert I want to pass the object to the client using json_encode(). The problem is, the _id value is not included.

$widget = array('text' => 'Some text');

$this->mongo->db->insert($widget);


If I echo $widget['_id'] the string value gets displays on the screen, but I want to do something like this:

$widget['widgetId'] = $widget['_id']->id;


So I can do json_encode() and include the widget id:

echo json_encode($widget);

Churchless

Answer 1

Believe this is what you're after.

$widget['_id']->{'$id'};

Something like this.

$widget = array('text' => 'Some text');
$this->mongo->db->insert($widget);
$widget['widgetId'] = $widget['_id']->{'$id'};
echo json_encode($widget);

Melodrama

Reference here: php.net/manual/en/class.mongoid.php. I'd prefer the (string) typecast below myself but at the time of the question I was using the method outlined in the docs. – Melodrama

Answer 2

You can also use:

(string)$widget['_id']

Jaguarundi

Answer 3

correct way is use ObjectId from MongoDB:

function getMongodbIDString($objectId){
    $objectId = new \MongoDB\BSON\ObjectId($objectId);
    return $objectId->jsonSerialize()['$oid'];
}

and do not cast the objectId like (string) $row['_id'] or $row->_id->{'$oid'}

Pandit

thanks for giving this code. no other code works except this...once thanks alot. – Heddi

@Dhanunjaya, nice, try use Doctrine ODM, it handle all this stuff – Chema

Answer 4

I used something similar:

(string)$widget->_id

Maryannemarybella

Thanks a lot :) it is working fine for me... $filter=['_id' => $row->_id]; and $filter you can pass to deleteOne and findOne function – Aubrette

Answer 5

I used something similar if object:

$widget->_id->{'$oid'}

or

(string)$widget->_id

or array :

$widget['id']->{'$oid'}
(string)$widget['_id']

Faustofaustus