How do I query for an object vs an array of objects in a document?
I need to select all the documents that are like Dataset 2.
Dataset 1:
{
'firstname' : 'John',
'lastname': 'Smith',
'assistance': [
{'type': 'Food', 'amount': 20}
]
}
Dataset 2:
{
'firstname' : 'John',
'lastname': 'Smith',
'assistance': {
'type': 'Food',
'amount': 20
}
}
you can use $type , $type won't work for dataset1, the reason being that it doesn't check if the field is array or not, what it checks is that the field contains array or not.
But if you are looking for dataset 2, you can use $type for object
db.whatever.find( { "assistance" : { $type : 3 } } );
or
db.whatever.find( { "assistance" : { $type : "object" } } );
MongoDB stores data in BSON format.
BSON stands for Binary JSON.
BSON supports various datatypes for values in documents.
According to documentation of MongoDB
$type selects the documents where the value of the field is an instance of the specified BSON type.
According to above mentioned documents
Dataset 1:Datatype of assistance field is an array.
Dataset2 : Datatype of assistance field is an object.
To retrieve only documents containing assistance key having object as a data type please try executing following query in MongoDB shell.
db.collection.find({assistance:{$type:3}})
For more detailed description regarding BSON types please refer documentation as mentioned in below URL
db.foo.find( { "assistance" : { $type : 3 } } ); // 3 is bson type for object
will return both the documents and
db.foo.find( { "assistance" : { $type : 4 } } ); // 4 is bson type for object
will return none of the two documents.
This is becuase in mongodb, when querying on an array, the array elements are checked instead on the complete array.
You can eliminate all elements where assistance is of type array by:
db.foo.find({"assistance.0" : {$exists : false}})