我正在使用Django渲染项目菜单。我可以得到我想要的内容,但是当我将其返回到django模板时,它只是一个json字符串,因此不会正确迭代。如何告诉django将其作为可迭代对象返回?
我遇到了this article。也许我想合并查询集?
Python
def index(request):
if not request.user.is_authenticated:
return redirect('/login', {'message': None})
try:
menu_categories = MenuCategory.objects.all()
menu = []
for cat in menu_categories:
items = MenuCategoryItems.objects.filter(category_id=cat.id).all()
menu.append({'category': cat, 'items': items})
context = {'menu': menu}
# for cat in menu_categories:
# items = menu_items.filter(category_id=cat.id)
# category_items = []
# for item in items:
# category_items.append({
# "name": item.name,
# "price": float(item.price),
# "id": item.id
# })
# menu.append({"category": cat.name, "items": category_items})
except Exception:
print('failure')
return render(request, 'index.html', context)
模板
{% for category in menu %}
<div>{{ category.name }}</div>
{# {% for item in category.items %}#}
{# <div>{{ item.name }} | {{ item.price }}</div>#}
{# {% endfor %}#}
{% endfor %}
答案 0 :(得分:0)
我想我找到了答案。这对我有用,但是我不确定这是否是“首选”方式。
Python
def index(request):
if not request.user.is_authenticated:
return redirect('/login', {'message': None})
try:
menu_categories = MenuCategory.objects.all()
menu = []
for cat in menu_categories:
items = MenuCategoryItems.objects.filter(category_id=cat.id).all()
menu.append({'category': cat, 'items': items.values})
context = {'menu': menu}
except Exception:
print('failure')
return render(request, 'index.html', context)
模板
{% for category in menu %}
<div><strong>{{ category.category.name }}</strong></div>
{% for item in category.items %}
<div>{{ item.name }} - {{ item.price }}</div>
{% endfor %}
{% endfor %}