Получение данных из трех таблиц (JOIN)
1- В таблице CategoryManager мы хотим отфильтровать по полю Manager в соответствии с пользователем запроса (например, result = category_manager).
2- Используя список category_manager, полученный на шаге 1, мы хотим перечислить все курсы, принадлежащие этим категориям, из таблицы Course.
class Category(models.Model):
id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False)
title = models.CharField(max_length=200)
class CategoryManager(models.Model):
category = models.ForeignKey(Category, on_delete=models.PROTECT, verbose_name=_('Course Category'), related_name="category_categorymanager")
manager = models.ForeignKey(_user, verbose_name=_('Category Manager'), on_delete=models.CASCADE, related_name="user_categorymanager")
class Course(models.Model):
id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False)
title = models.CharField(max_length=200, verbose_name=_('Course Name'))
...
price = models.DecimalField(max_digits=7, decimal_places=2, verbose_name=_('Course Price'), blank=True, null=True)
slug = models.SlugField(unique=True, blank=True)
category = models.ForeignKey(Category, related_name='courses_category', on_delete=models.PROTECT)
Вы можете фильтровать с помощью:
@admin.register(Course)
class CourseAdmin(admin.ModelAdmin):
def get_queryset(self, request):
return Course.objects.filter(
category__category_categorymanager__manager=request.user
).distinct()
Note: Your
CategoryManager
model acts as a junction table for a many-to-many relation betweenCategory
and_user
. You can span aManyToManyField
[Django-doc] on theCategory
model with:class Category(models.Model): # … managers = models.ManyToManyField( _user, through='CategoryManager' )
Note: It is normally better to make use of the
settings.AUTH_USER_MODEL
[Django-doc] to refer to the user model, than to use theUser
model [Django-doc] directly. For more information you can see the referencing theUser
model section of the documentation.