I have a document in MongoDB and I'm trying to unwind it in PHP. I want to unwind a document that has a subdocument that contains yet anothersubdocument. I was able to do this succesfully if the document only contains strings and numbers, but if it contains another subdocument then I can't get it to work. I get this error:
exception: $unwind: value at end of field path must be an array
Can you not unwind a subdocument that contains another level of subdocument? If not, how would you go about doing this?
Thanks in advance!
This is the query:
$project = array(
                '$project' => array(
                    '_id' => 1,
                    'items' => 1,
                )
            );
$unwind = array(
                '$unwind' => '$items'
            );
 $query = $mongo->store->aggregate($project,$unwind_items);
This is the structure:
{
    "_id": {
        "$oid": "526fdc1fd6b0a8182300009c"
    },
    "items": [
               {
            "quantity": "1",
            "category_id": {
                "$oid": "526fdc1fd6b0a81823000029"
            },
            "category": "test",
            "images": [
                {
                    "name": "9by9easy.PNG",
                    "path": "upload_files/nibh-vulputate-mauris-corporation/",
                    "file_path": "upload_files/nibh-vulputate-mauris-corporation/68e7c50bde1476e96ca2461dc553cce5528fb70e41b1f.PNG",
                    "size": 8761
                },
                {
                    "name": "9by9hard.PNG",
                    "path": "upload_files/nibh-vulputate-mauris-corporation/",
                    "file_path": "upload_files/nibh-vulputate-mauris-corporation/8cd2dcf4fcd476262db2eba3fdb2c39a528fb70e42757.PNG",
                    "size": 11506
                }
            ],
            "link": "fghfhfhfg"
        }
    ],
    "name": "Nibh Vulputate Mauris Corporation",
}
                To unwind, use $unwind. The $unwind deconstructs an array field from the input documents to output a document for each element.
Definition. $unwind. Deconstructs an array field from the input documents to output a document for each element. Each output document is the input document with the value of the array field replaced by the element.
Accessing embedded/nested documents – In MongoDB, you can access the fields of nested/embedded documents of the collection using dot notation and when you are using dot notation, then the field and the nested field must be inside the quotation marks.
The $project takes a document that can specify the inclusion of fields, the suppression of the _id field, the addition of new fields, and the resetting of the values of existing fields. Alternatively, you may specify the exclusion of fields. Specifies the inclusion of a field.
I know that what you're trying to do is possible with MongoDB. The aggregate() command can take as many arguments as you need.
In the mongo shell, a command like this
db.collection.aggregate(
  { $project: {
    _id: 1,
    items: 1
  } },
  { $unwind: '$items' },
  { $unwind: '$items.images' }
);
will unwind the items subdocument, then the images subdocument. 
Based on the code in your question, perhaps this will work
$project = array(
  '$project' => array(
    '_id' => 1,
    'items' => 1,
  )
);
$unwind_items = array(
  '$unwind' => '$items'
);
$unwind_images = array(
  '$unwind' => '$items.images'
);
$query = $mongo->store->aggregate($project,$unwind_items,$unwind_images);
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