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);

score 45 · Accepted Answer · answered Oct 28 '11 at 06:36

45

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);

answered Oct 28 '11 at 06:36

John Pancoast

1,241
16
14

2

Reference here: http://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. – John Pancoast Nov 18 '14 at 21:27

score 23 · Answer 2 · answered Jan 14 '13 at 20:16

23

You can also use:

(string)$widget['_id']

answered Jan 14 '13 at 20:16

meltix

311
3
4

score 5 · Answer 3 · answered Sep 25 '19 at 18:04

5

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'}

answered Sep 25 '19 at 18:04

Rafael Guimarães

400
4
10

thanks for giving this code. no other code works except this...once thanks alot. – Dhanunjaya Oct 30 '19 at 07:55
@Dhanunjaya, nice, try use Doctrine ODM, it handle all this stuff – Rafael Guimarães Oct 30 '19 at 17:08

score 2 · Answer 4 · edited Mar 26 '18 at 08:08

2

I used something similar:

(string)$widget->_id

edited Mar 26 '18 at 08:08

Ikhlak S.

8,578
10
57
77

answered Jul 09 '13 at 11:59

Abilash

21
6

Thanks a lot :) it is working fine for me... $filter=['_id' => $row->_id]; and $filter you can pass to deleteOne and findOne function – Venkatgreat309 Mar 11 '20 at 14:51

score 1 · Answer 5 · answered Feb 06 '18 at 09:04

1

I used something similar if object:

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

or

(string)$widget->_id

or array :

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

answered Feb 06 '18 at 09:04

Mahdi Bagheri Varnosfatherani

97
1
9

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

5 Answers5

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