What default id should I use such that all the ids in my collection are greater than this id? - mongodb

Question

I intend to fetch 100 ids at once in a sorted manner.
I find the ids greater than skip where skip can be set to a default value at the beginning. I need to sort the ids generated in the find() and the limit set is 100.
So, my query is:
db['Organization'].find({"_id":{"$gt":ObjectId(skip)}},{"_id":1}).sort([("_id",1)]).limit(100)
As of now, I have set skip to str(0). I intend to update it with the last id fetched in the iteration.
The complete endpoint is:
#hug.get('/organization/collect_pricing')
def get_organizations():
start_time = datetime.strptime('2016-11-01', '%Y-%m-%d')
org_ids = []
org_pricing_plans = []
counter = 0
skip = str(0)
result_check = True
pricing_response = []
ob_toolbox = Toolbox()
while(result_check is True):
print(counter)
try:
if
organizations = db['Organization'].find({"_id":{"$gt":ObjectId(skip)}},{"_id":1}).sort([("_id",1)]).limit(100)
except Exception as e:
print(e)
if organizations.count(True) == 0:
result_check = False
continue
counter += 100
for org in organizations:
org_ids.append("Organization$"+org["_id"])
try:
pricing_plans = ob_toolbox.bulk_fetch(collection="PricingPlan", identifier="_p_organization", ids=org_ids)
except Exception as e:
print(e)
currDict = {}
for i in range(0, organizations.count(True)):
currDict["id"] = org_ids[i]
currDict["expiresAt"] = pricing_plans[i]["expiresAt"]
currDict["resources"] = pricing_plans[i]["resources"]
currDict["_created_at"] = pricing_plans[i]["_created_at"]
org_pricing_plans.append(currDict)
print(currDict["id"])
skip = currDict["id"]
if organizations.count(True) < 100:
result_check = False
return (org_pricing_plans)

Answer 1

If you want the default "minimal" value, then null object id is better. It's the same type (ObjectId) and will sort lowest.
ObjectId('000000000000000000000000')
Alternatively, you could branch when doing a query. Is it first query? If yes, don't include the skip part. If no, use last id from previous results.