Question

我有这两个型号。我需要返回每个部门最老的员工名单。

class Employee(models.Model):
    name = models.CharField(max_length=50)
    birth_day = models.DateField()
    department = models.ForeignKey('Department')

class Department(models.Model):
    sector = models.CharField(max_length=255)

我有点担心如何在Django中制作它。

Answer 1

我认为没有直接查询。您需要建立此类员工的名单。

empList = []
for dept in Department.objects.all() :
    empList = empList + [dept.employee_set.order_by('birth_day')[0]]

#empList will have list oldest of employees.

Answer 2

我认为这个决心。

empList = []
for dept in Department.objects.all() :
    empList.append(dept.employee_set.latest('birth_day'))

但我也对更好的方法感兴趣，我正在想办法只做一个查询。

我试过了：

In [98]: d = Employee.objects.filter().annotate(Min('birth_day'))

In [99]: d.group_by = ['department_id']

In [100]: x = QuerySet(query=d, model=Employee)
(0.001) QUERY = 'SELECT "permissoes_employee"."id", "permissoes_employee"."name", "permissoes_employee"."birth_day", "permissoes_employee"."department_id", MIN("permissoes_employee"."birth_day") AS "birth_day__min" FROM "permissoes_employee" GROUP BY "permissoes_employee"."id", "permissoes_employee"."name", "permissoes_employee"."birth_day", "permissoes_employee"."department_id"' - PARAMS = (); args=()

但它不是仅仅由部门进行分组，而是忽略＆＃39; group_by并以对所有字段进行分组结束，最终会产生错误的结果。

我认为我们没有比这更好的解决方案，只是为了使cursor.execute（＆＃34; SQL＆＃34;）

：（

列出每个部门的最老员工

2 个答案: