两种型号：

class University(models.Model):
    name = models.CharField("name", max_length = 48)
    city = models.ForeignKey(City, on_delete = models.CASCADE) 

class Course(models.Model):
    name = models.CharField("course name", max_length = 48)
    university = models.ForeignKey(University, on_delete = models.CASCADE)
    students = models.ManyToManyField(Student, related_name = "%(class)s_name", related_query_name = "student_course_qs", blank = True)
    online = models.BooleanField("online course", default = False)
    semester = models.PositiveIntegerField("choose 1-6", default = 1, choices = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)])
    time_update = models.DateTimeField("last info update", auto_now = True)

限制条件：如果Course具有students，则它不能是online，反之亦然。
如何获得给定city的所有Universities上的所有Courses的列表，其中Course上的最新条目包含students，而Course包含在第三个semester中？
我尝试过在University的不同组上循环，但这会导致大量的查询：

universities = University.objects.filter(city = City.objects.first())

wanted = []
for univ in universities:
    c = Course.objects.filter(university = univ).order_by("time_update").last()
    if c.semester == 3 and not c.online:
        wanted.append(c.id)

selected_courses = Courses.objects.filter(id__in = wanted)

我在这里看到的问题是，我创建了太多的查询，并且在universities的过滤器之后，它们就已经被分组了-我可以一个接一个地处理这些组而不重新创建查询吗？