How to keep orders in MongoDB? - mongodb

Question

In my MongoDB document I have object like this
[_id] => MongoId Object (
[$id] => 52a46b44aabacb5c218b4567
)
[results] => Array (
[http://google.com] => Array (
[position] => 1
[data] => 42672
)
[http://bing.com] => Array (
[position] => 2
[data] => 9423
)
[http://yandex.com] => Array (
[position] => 3
[data] => 5513
)
)
I would like to change data parameter in "bing.com" from 9423 to for instance 300. Moreover, I have to keep order of the sites. It have to looks like this
[_id] => MongoId Object (
[$id] => 52a46b44aabacb5c218b4567
)
[results] => Array (
[http://google.com] => Array (
[position] => 1
[data] => 42672
)
[http://bing.com] => Array (
[position] => 2
[data] => 300
)
[http://yandex.com] => Array (
[position] => 3
[data] => 5513
)
)
Is this achievable in Mongo?

Answer 1

The reordering of fields issue has been fixed as of MongoDB v2.5.2 (2.6 release). Having said that one way you can avoid the issue completely is having results as an array instead of a (sub)document. Also note you should not use "." as part of the key name either.
With 2.4, with the following you will see there is reodering in the case of _id=1 (subdocument) but not in the case of _id=2 (array).
$document = array("_id" => 1, "results" => array('http://google.com' => array('position' => 1, 'data' => 42672),
'http://bing.com' => array('position' => 2, 'data' => 9423),
'http://yandex.com' => array('position' => 3, 'data' => 5513)));
$coll->insert($document);
$document = array("_id" => 2, "results" => array(array('site' => 'http://google.com', 'data' => 42672),
array('site' => 'http://bing.com', 'data' => 9423),
array('site' => 'http://yandex.com', 'data' => 5513)));
$coll->insert($document);
$coll->update(array("_id" => 1), array('$set'=>array("results.http://bing.com.data"=>300)));
$coll->update(array("_id" => 2, 'results.site' => 'http://bing.com'), array('$set'=>array('results.$.data'=>300)));

Answer 2

I've included examples below using the mongo shell for clarity, but the PHP equivalent should be straightforward to work out.
I notice you originally modelled your list of sites as an embedded document, however the order of fields within an embedded document is currently not guaranteed to be preserved so you should instead use an array.
Additionally, you cannot use field names with embedded dots (.) in MongoDB so you should not plan to store urls as field names (see: Field name restrictions).
In order to find an element in an array you need to search by a value (not a field name) so your schema should look more like:
{
_id: ObjectId("52a46b44aabacb5c218b4567"),
results: [
{
site: 'http://google.com',
position: 1,
data: 42762
},
{
site: 'http://bing.com',
position: 2,
data: 9423
},
{
site: 'http://yandex.com',
position: 3,
data: 5513
}
]
}
Assuming the array site elements are unique, you can use the positional operator $ to find and update the matching embedded document in place.
For example, to perform your update of the "bing.com" data value:
db.sites.update(
// Match criteria
{
_id:ObjectId("52a46b44aabacb5c218b4567"),
'results.site':'http://bing.com'
},
// Update
{ $set: {
'results.$.data': 300 }
}
)
In MongoDB 2.4+ you have the option of pushing to a sorted array which could also be a useful approach to maintaining your array in sorted order when you add new entries.
It's worth noting that if you plan to store many (i.e. thousands) of items in an array this can impose a significant performance penalty due to document growth and the complexity of updating large arrays.

Answer 3

I am pretty sure that (as every other DBMS) you can't and should't rely on records orders.
Instead I would advice you to add index (on position, i.e. db.people.ensureIndex( { position: 1 } )) and query your record sorted by that field, i. e.: db.collection.find().sort( { position: 1 } )