我应该使用哪个默认ID,以便集合中的所有ID都大于此ID?

我打算按顺序一次取100个身份证。 我 find ID大于 skip 其中skip可以在开始时设置为默认值。我需要对 find() 限制是100。 所以,我的问题是:

db['Organization'].find({"_id":{"$gt":ObjectId(skip)}},{"_id":1}).sort([("_id",1)]).limit(100)

到目前为止,我已经将skip设置为str(0)。我打算用迭代中获取的最后一个id更新它。

完整的端点是:

@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)