Parse Error in PHP with MongoDB

Question

I'm trying to query a mongodb database with the _id stored in a session variable...

getting the _id and store as session variable:

$_SESSION['user']['userid']=$user['_id']->{'$id'};

querying db - error on this line:

$user=collection->findOne(array('_id' => new MongoId($_SESSION['user']['userid'])));

can't figure it out.

Shankar Narayana Damodaran · Accepted Answer · 2014-02-27 11:22:11Z

2

Should be $user=$collection->findOne(array('_id' => new MongoId($_SESSION['user']['userid']))); You missed a $ infront of collection

answered Feb 27, 2014 at 11:22

community wiki

Sign up to request clarification or add additional context in comments.

4 Comments

Now I get a Notice: array to string conversion

Are you getting that on your echo statement ?

Replace echo with print_r. That is because an array can't be echoed all together.

You need to json_encode the array and then pass it off.

Vikram Jain · Accepted Answer · 2014-02-27 11:24:43Z

0

<?php

$users= $mongo->my_db->users;
$user = $users->findOne(array('_id' => new MongoId($_SESSION['user']['userid']));
print_r($user);

?>

answered Feb 27, 2014 at 11:24

Vikram Jain

5,6081 gold badge22 silver badges31 bronze badges