假设我有两个不同的应用程序:teacher/models.py: Teacher(models.Model): name = models.CharField(max_length=300)class/models.py: Class(models.Model): name = models.CharField(max_length=300) teacher = models.ForeignKey(Teacher) students = models.ManyToManyField(Student)我想让所有的老师上课和所有的课程都附上。我想要的结果:{[ teacher: '3L' #Teachers Id classes: ['20L','14L','30L'] #list of Class objects or ids with the above teacher],[# similar to above]}这是可能的吗?这就是我目前正在做的事情:classes = Class.objects.all()teachers = Teacher.objects.filter(id__in=classes.value_list('teacher',flat=True).distinct())for teacher in teachers: classes_for_teachers = classes.objects.filter(teacher=teacher)在上面的代码中,有四个使用循环进行的查询,这无疑增加了时间复杂度。有没有更好的解决方案?提前致谢。
1 回答

猛跑小猪
TA贡献1858条经验 获得超8个赞
使用prefetch_related:
teachers = Teacher.objects.prefetch_related('class_set')
# what you want is not a valid Python structure (set of lists (looking like dicts))
# list of dicts makes more sense
result = [
{'teacher': t.pk, 'classes': t.class_set.all()}
for t in teachers
]
无论有多少老师和班级,这只会触发 2 db 查询。
没有找到匹配的内容?试试慕课网站内搜索吧
添加回答
举报
0/150
提交
取消