Question

我正在使用django，我必须恢复名为Restaurante的模型的记录，该模型具有名称，地址，电话，城市等属性，并且具有food_type属性，该属性是ManyToManyFiel，使用我称为FoodType的模型，因为在餐厅中您可以提供多种类型的食物（例如墨西哥，意大利，素食主义者等）和某种类型的食物，因此可以在许多餐厅中提供它，并且我想恢复所有餐厅在X城市中，并按照以下食物类型将其分组：

素食食品
- 堆肥机
- 天然食品
- GreenRestaurant
- Resto Bar
意大利美食
- Resto Bar
- 派披萨
- 堆肥机
墨西哥食物
- Jalisquitos
- 玛雅炸玉米饼
- GreenRestaurant

我尝试使用标签{％regroup％}来完成此操作，但是它显示了预期的输出，我附上了模型代码：

class FoodType(models.Model):
    name = models.CharField(
        max_length = 150,
        blank = False,
        null = False,
        help_text = 'Enter the name of the type of food',
        verbose_name = 'Name'
    )
    description = models.CharField (
        max_length = 250,
        blank = False,
        null = False,
        help_text = 'Enter a brief description of this type of food',
        verbose_name = 'Description'
    )

class Restaurant(models.Model):
    name = models.CharField (
        max_length = 50,
        blank = False,
        null = False,
        help_text = 'Enter the name of the restaurant',
        verbose_name = 'Restaurant Name',
    )
    address = models.CharField (
        max_length = 200,
        blank = False,
        null = True,
        help_text = 'Enter the physical address of the restaurant',
        verbose_name = 'Addres'
    )
    telefono = PhoneNumberField(
        null = False,
        help_text = 'Enter the phone number of the restaurant in the format +999999999',
        unique = True,
        verbose_name = 'Phone Number'
    )
    city = models.ForeignKey(
        City,
        related_name = 'ubica_en',
        on_delete = models.DO_NOTHING,
        blank = False,
        null = True,
        help_text = 'Select the city to which this restaurant belongs',
        verbose_name = 'City',
    )
    administrator = models.ForeignKey(
        User,
        related_name = 'es_registrado_por',
        on_delete = models.CASCADE,
        blank = False,
        null = False,
        help_text = 'User who added this restaurant',
        verbose_name = 'Who registered it?'
    )
    food_type = models.ManyToManyField(
        FoodType,
        help_text = 'Select the type (s) of food served in this restaurant',
        related_name = 'restaurants',
        verbose_name = 'Type of food',
        blank = False
    )

我在代码中附加了标签{％regroup％}：

{% regroup restaurants by food_type as food_type_list %}
{% for food_type in food_type_list %}
  {{ food_type.grouper }}
  {% for restaurant in food_type.list %}
    <h1>{{ restaurant|upper }}</h1>
  {% endfor %}
{% endfor %}

希望您能提前帮助我，非常感谢。

Answer 1

最后，我通过以下方式解决了该问题：

foods = FoodType.objects.all()
group = {}
for food in foods:
    restaurant_key = '_'.join([str(restaurant.pk) for restaurant in comida.restaurant_set.all()])
    if not restaurant_key in group and restaurant_key! = '':
        group[restaurant_key] = {'restaurants': food.restaurant_set.all(), 'foods': []}
    if restaurant_key! = '':
        group[restaurant_key]['foods'].append(food)
context ['group'] = group

通过ManyToManyField重新组合查询集

1 个答案: