查询中的Django绝对值

如果愿意,可以使用exclude将不匹配查询的对象提供给您

#this returns true if one or more and dont need to use len()
#if it looks better for you you can try it out
#if everything is fine will return None
if User.objects.filter().exclude(user__in=uiids).first(): 
   raise your_error

别想得太复杂。您想知道某个列表是否完全包含在另一个列表中,这只是简单的python:

user_ids = User.objects.values_list('user_id',flat=True)
if not set(uuids).issubset(user_ids): # or use .difference() to get the uuids missing from user_ids
    raise stuff