我想在不使用mptt或django-tree的情况下为餐厅菜单建立结构数据库。这是我的models.py:
from django.db import models
class Menu(models.Model):
name = models.CharField(max_length=24, unique=True, verbose_name='menu name')
#slug = models.SlugField(max_length=24, unique=True, help_text='The slug is the URL friendly version of the menu name, so that this can be accessed at a URL like mysite.com/menus/dinner/.')
additional_text = models.CharField(max_length=128, null=True, blank=True, help_text='Any additional text that the menu might need, i.e. Served between 11:00am and 4:00pm.')
order = models.PositiveSmallIntegerField(default=0, help_text='The order of the menu determines where this menu appears alongside other menus.')
class Meta:
ordering = ['name', 'order']
class MenuCategory(models.Model):
menu = models.ForeignKey(Menu, on_delete=models.CASCADE,help_text='The menus that this category belongs to, i.e. \'Lunch\'.')
name = models.CharField(max_length=32, verbose_name='menu category name')
additional_text = models.CharField(max_length=128, null=True, blank=True, help_text='The additional text is any bit of related information to go along with a menu category, i.e. the \'Pasta\' category might have details that say \'All entrees come with salad and bread\'.')
order = models.IntegerField(default=0, help_text='The order is the order that this category should appear in when rendered on the templates.')
class Meta:
verbose_name='menu category'
verbose_name_plural='menu categories'
ordering = ['order', 'name']
def __unicode__(self):
return self.name
class MenuItem(models.Model):
CLASSIFICATION_CHOICES = (
('neither', 'Neither'),
('vegan', 'Vegan'),
('vegetarian', 'Vegetarian'),
)
name = models.CharField(max_length=48, help_text='Name of the item on the menu.')
description = models.CharField(max_length=128, null=True, blank=True, help_text='The description is a simple text description of the menu item.')
category = models.ManyToManyField(MenuCategory, verbose_name='menu category', help_text='Category is the menu category that this menu item belongs to, i.e. \'Appetizers\'.')
order = models.IntegerField(default=0, verbose_name='order', help_text='The order is to specify the order in which items show up on the menu.')
price = models.IntegerField(help_text='The price is the cost of the item.')
image = models.ImageField(upload_to='menu', null=True, blank=True, verbose_name='image', help_text='The image is an optional field that is associated with each menu item.')
class Meta:
verbose_name='menu item'
verbose_name_plural='menu items'
ordering = ['classification', 'order', 'name']
def __unicode__(self):
return self.name
views.py:
from .models import Menu,MenuCategory
from django.views.generic import ListView
class MenuView(ListView):
model= Menu
conext_object_name='name'
template_name = 'menu_list.html'
queryset = Menu.objects.all()
def get_context_data(self, **kwargs):
context = super(MenuView, self).get_context_data(**kwargs)
context['Menucategory'] = MenuCategory.objects.all()
context['Menus'] = self.queryset
return context
menu_list.html:
<div>
{%for item in Menu%}
<p>
{{item.menu}}
{{item.name}}
{{item.additiona_text}}
{{item.order}}
</p>
{%endfor%}
{%for item in MenuCategory%}
<p>
{{item.name}}
{{item.additiona_text}}
{{item.order}}
</p>
{%endfor%}
</div>
当我执行runserver,后浏览本地主机页面时,它仅显示
Django博客
早餐1
午餐2
晚餐3
但是我想要的输出应该是:
1]早餐
咖啡
物品
1 2 3
零食1 2 3
2]午餐
入门项目1 2 3
主要课程项目1 2 3
我如何使用普通的引导表行列获取此结构化表,在每行列中我都会使用for循环查询项目?他们还有其他办法吗?如果有其他方法,我也有兴趣。...
我认为您的设计一开始就不好。如果您想以这种方式显示事物,则逻辑如下:
因此,您的数据结构应为以下内容:
:用其类名命名查询集是一种非常不好的做法。只需将其命名为menus!这就是我要做的此处),并带有一个双class MenuItem(models.Model): # [...] class MenuCategory(models.Model): # [...] name = CharField... items = models.ManyToManyField(MenuItem) class Menu(models.Model): # [...] menu_category = models.ForeignKey(MenuCategory)
然后,您可以在模板中做您想做的事(note
for
循环:<div> {% for menu in menus %} <h3> {{ menu.menu_category.name }} </h3> <p> {%for item in menu.menu_category.items.all %} {{ item.name }} {% endfor %} </p> {% endfor %} </div>
再次,您的初始设计在我看来不是最合适的。
希望有帮助!