Python Django：如何以非恐怖的方式查询child、granchildren等记录

我试图返回Employees表中与用户具有嵌套关系的用户列表/筛选器。例如，我将员工绑定到他们的经理，我希望能够查询该经理下的所有员工（这包括主经理下的任何其他经理下的任何员工）。因此，如果用户

Bob

有两个直接下属，

Sally

和

Brian

。而

Brian

有2名直接下属，

Sally

有3名直接下属。我希望Bob能够看到所有7名员工。现在，我能让它工作的唯一方法是通过一个可怕的序列，如下所示..我希望他们的方法更简单/更有效

    manager = Employees.objects.filter(manager_id=request.user.id).values('manager')
    employee_ids = list(Employees.objects.filter(manager=manager.first()['manager']).values_list('employee', flat=True))
    employees = [User.objects.get(id=i).username for i in employee_ids]
    grandchildren = []
    for i in employees:
        user_id = User.objects.get(username=i).id
        child = list(Employees.objects.filter(manager=user_id).values_list('employee', flat=True))
        grandchildren.append(child)
    children = list(chain.from_iterable(grandchildren))
    for i in children:
        user_id = User.objects.get(id=i).id
        child = list(Employees.objects.filter(manager=user_id).values_list('employee', flat=True))
        grandchildren.append(child)
    grandchildren = list(chain.from_iterable(grandchildren))
    for i in grandchildren:
        employees.append(User.objects.get(id=i).username)
    employees = list(set(employees))

---

对不起，你的代码看起来很糟糕。首先，我指的是太多的DB查询（大多数查询都是非优化的，甚至不需要）

根据您的描述，我建议您尝试以下方式：

manager_id = request.user.id
children_ids = list(
    Employees.objects.filter(manager_id=manager_id).values_list('employee', flat=True)
)
grandchildren_ids = list(
    Employees.objects.filter(manager_id__in=children_ids).values_list('employee', flat=True)
)
# If you want to go deeper, do this in a loop and stop once an empty list of IDs is fetched 
# (which means that there are no descendants anymore)

# Combine all IDs and finally fetch the actual users 
# (do it only once, and fetch all the users in a single query, not one by one)
employees_ids = children_ids + grandchildren_ids
employees = User.objects.filter(id__in=employees_ids)

旁白：这是一个笑话吗

user\u id=user.objects.get（id=i.id

？：）