I dont know why i cant use aggregation

Question

Hi,

I was wondering why my queries are not working when i am using aggregation in cosmos.

from azure.cosmos import CosmosClient, PartitionKey

cosmos_url = 'https://xxx.documents.azure.com:443/'
cosmos_key = 'xxx=='
client = CosmosClient(cosmos_url, cosmos_key)

# Define your Cosmos database and container
database_name = 'xxxx'
container_name = 'xxxx'
database = client.get_database_client(database_name)
container = database.get_container_client(container_name)

cosmos_query = "SELECT count(c.x) AS x FROM c WHERE (c.x= 'x')"

query_result = list(container.query_items(
    query=cosmos_query,
    enable_cross_partition_query=True 
))
for item in query_result:
    print(item)

Its giving me error as below :-

azure.cosmos.exceptions.CosmosHttpResponseError: (BadRequest) {"code":"BadRequest","message":"Query contains the following features, which the calling client does not support:\nNone NonValueAggregate \r\nActivityId:xxx, Windows/10.0.20348 cosmos-netstandard-sdk/3.18.0"} 

While i am running the same query at the portal i am getting a result.

I am using azure-cosmos 4.5.1

and python 3.9

Please help

Answer 1

Hi Sakib,

Is possible can you try upgrading Phython SDK library that supports the value? - https://learn.microsoft.com/en-us/azure/cosmos-db/nosql/sdk-python - this helped most of the users as per my experience.

you also try alternate solution.

cosmos_query = "SELECT c FROM c WHERE c.x = 'x'"

query_result = list(container.query_items(

query=cosmos_query,

enable_cross_partition_query=True

))

count = len(query_result) # Count the number of returned documents

print(f"Count: {count}")

Note- enable_cross_partition_query=True is only needed if your query need to access data from multipel partitions.

Please accept if this answer if helped, do let me know if you have additional question.

Answer 2

Hi @VasimTamboli is there any other solution.

Im facing the same problem, and the alternative could serve for a small amount of data, but it not effective for a big one..

@Anonymous @Sakib Ali Choudhary did you find any other solution?