我有一个集合,我想将其转换为list以便使用。我就是做不到。
@tasks.loop(minutes=1)
async def check_current_mutes(self):
print("AAAAAA")
mutes = utils.servicecode.IslaDB("Isla", "Muted")
print(datetime.datetime.now())
lists = mutes.test_collection.find({'unmute': {'$gte': str(datetime.datetime.now())}})
print(lists)
for muted in lists.to_list(length=100):
print(muted)
if muted["unmute"] == "Indefinite":
return
else:
if not time.strptime(muted["unmute"]) <= str(datetime.datetime.now()):
member = await self.bot.get_user(muted)
print(member)我只需要迭代它来比较它中的值。
发布于 2022-09-05 03:59:26
我认为问题是,您必须等待将该对象转换为列表如下:
...
lists = mutes.test_collection.find({'unmute': {'$gte': str(datetime.datetime.now())}})
my_data_as_list = await lists.to_list(length=100)
for muted in my_data_as_list:
print(muted)
...这是你错过的一步:
my_data_as_list = await lists.to_list(length=100)https://stackoverflow.com/questions/67613910
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