get_queryset中的Django 2.0 url参数 [英] Django 2.0 url parameters in get_queryset
问题描述
我想根据网址中的类别ID过滤子类别
对于恒定值,它可以正常工作
return Subcategory.objects.filter(category = 1)
views.py
class SubcategoriesListView(ListView):
模型=子类别
template_name ='app / categories / index.html'
def get_queryset(self):
返回Subcategory.objects.filter(category = category_id)
urls.py
path('categories /< int: category_id> /',app.views.SubcategoriesListView.as_view(),name ='subcategories'),
models.py
class子类别(models.Model):
title = models.CharField(max_length = 30)
category = models.ForeignKey(Category,on_delete = models.CASCADE)
Traceback p>
/ catego处的NameError ries / 1 /
名称'category_id'未定义
get_queryset中的views.py
返回Subcategory.objects.filter(category = category_id)
您可以使用 self.args在基于类的视图中获取URI位置和命名参数。
(元组)和 self.kwargs
(字典)。
此处您将 category_id
定义为命名参数,因此可以使用 self.kwargs ['category_id']
:
class SubcategoriesListView(ListView):
model =子类别
template_name ='app / categories / index.html'
def get_queryset(self):
返回Subcategory.objects.filter( category_id = self.kwargs ['category_id'] )
由于 id
是整数,因此您对 category_id
,不在 category
上。
I would like to filter subcategories based on the category id from the url
For a constant value it works without a problem
return Subcategory.objects.filter(category = 1)
views.py
class SubcategoriesListView(ListView):
model = Subcategory
template_name = 'app/categories/index.html'
def get_queryset(self):
return Subcategory.objects.filter(category = category_id)
urls.py
path('categories/<int:category_id>/', app.views.SubcategoriesListView.as_view(), name='subcategories'),
models.py
class Subcategory(models.Model):
title = models.CharField(max_length=30)
category = models.ForeignKey(Category, on_delete=models.CASCADE)
Traceback
NameError at /categories/1/ name 'category_id' is not defined
views.py in get_queryset return Subcategory.objects.filter(category = category_id)
You can obtain the URI positional and named parameters in a class-based view with self.args
(a tuple) and self.kwargs
(a dictionary) respectively.
Here you defined the category_id
as a named parameter, so you can obtain its corresponding value with self.kwargs['category_id']
:
class SubcategoriesListView(ListView):
model = Subcategory
template_name = 'app/categories/index.html'
def get_queryset(self):
return Subcategory.objects.filter(category_id=self.kwargs['category_id'])
Since the id
is an integer, you thus filter on category_id
, not on category
.
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