如何使用Django-filter过滤多个字段？
如何使用Django-filter过滤多个字段？
提问于 2021-09-05 15:51:13
我想检索按用户和类别过滤的数据。
我正在使用Django Rest框架。
models.py
class Practice(models.Model):
practice_id = models.BigAutoField(primary_key=True)
user = models.ForeignKey('User', models.DO_NOTHING, default=None)
subcategory = models.ForeignKey('Subcategory', models.DO_NOTHING, default=None)
score = models.SmallIntegerField(null=True)
def __str__(self):
return str(self.practice_id)
class User(models.Model):
user_id = models.BigAutoField(primary_key=True)
fullname = models.CharField(max_length=50)
def __str__(self):
return self.fullname
class Subcategory(models.Model):
subcategory_id = models.BigAutoField(primary_key=True)
category = models.CharField(max_length=30)
def __str__(self):
return f"{ self.category }"serializers.py
class PracticeSerializer(serializers.ModelSerializer):
subcategory = SubcategorySerializer()
class Meta:
model = Practice
fields = ('practice_id',
'user',
'subcategory',
'score',
)views.py
@api_view(['GET'])
def practiceFilter_User(request):
if request.method == 'GET':
exercises = Practice.objects.all()
user = request.GET.get('user', None)
if user is not None:
practice_filtered = exercises.filter(user__user_id__icontains=user)
exercises_serializer = PracticeSerializer(practice_filtered, many=True)
return JsonResponse(exercises_serializer.data, safe=False)urls.py
urlpatterns = [
url(r'^api/practice-filter-user$', views.practiceFilter_User),
]在我的数据库中，我有以下3个实践数据：
[
{
practice_id: 1,
user: 1,
subcategory: {
subcategory_id: 1,
category: "Math",
},
score: 7,
},
{
practice_id: 2,
user: 1,
subcategory: {
subcategory_id: 1,
category: "Math",
},
score: 8,
},
{
practice_id: 3,
user: 1,
subcategory: {
subcategory_id: 2,
category: "History",
},
score: 9,
}
]如果执行上述代码是为了通过用户id = 1获取实践数据，则结果如下：
api/practice-filter-user?user=1
[
{
practice_id: 1,
user: 1,
subcategory: {
subcategory_id: 1,
category: "Math",
},
score: 7,
},
{
practice_id: 2,
user: 1,
subcategory: {
subcategory_id: 1,
category: "Math",
},
score: 8,
},
{
practice_id: 3,
user: 1,
subcategory: {
subcategory_id: 2,
category: "History",
},
score: 9,
}
]我希望检索所有id 1的user和所有Math category过滤的数据，如下所示：
api/practice-filter-subcategory-user?user=1&category=Math
[
{
practice_id: 1,
user: 1,
subcategory: {
subcategory_id: 1,
category: "Math",
},
score: 7,
},
{
practice_id: 2,
user: 1,
subcategory: {
subcategory_id: 1,
category: "Math",
},
score: 8,
}
]在检索用户和类别筛选的实践数据时，上述结果是我想要的。
我尝试在这个链接上使用django过滤器，但是出现了错误。
我向views.py添加了一个类
class practiceFilter_Subcategory_User(generics.ListAPIView):
queryset = Practice.objects.all()
serializer_class = PracticeSerializer
filter_fields = ('user', 'category')我在urls.py中添加了url
urlpatterns = [
url(r'^api/practice-filter-subcategory-user$', views.practiceFilter_Subcategory_User),
]我尝试用id 1和所有Math category检索由所有Math category过滤的数据，如下所示：
api/practice-filter-subcategory-user?user=1&category=Math
但是我失败了，并得到了上面的错误。
是否可以通过id 1和所有user过滤所有Math category？
更新我得到的跟踪错误:
Environment:
Request Method: GET
Request URL: http://192.168.1.2:8080/api/practice-filter-subcategory-user?user_id=1&category=Math
Django Version: 3.2.5
Python Version: 3.7.9
Installed Applications:
['django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
'rest_framework',
'rest_framework_bulk',
'yasn.apps.yasnConfig',
'corsheaders',
'django_filters']
Installed Middleware:
['django.middleware.security.SecurityMiddleware',
'django.contrib.sessions.middleware.SessionMiddleware',
'django.middleware.common.CommonMiddleware',
'django.middleware.csrf.CsrfViewMiddleware',
'django.contrib.auth.middleware.AuthenticationMiddleware',
'django.contrib.messages.middleware.MessageMiddleware',
'django.middleware.clickjacking.XFrameOptionsMiddleware',
'corsheaders.middleware.CorsMiddleware',
'django.middleware.common.CommonMiddleware']
Traceback (most recent call last):
File "C:\Users\Me\Documents\Python\env\lib\site-packages\django\core\handlers\exception.py", line 47, in inner
response = get_response(request)
File "C:\Users\Me\Documents\Python\env\lib\site-packages\django\core\handlers\base.py", line 181, in _get_response
response = wrapped_callback(request, *callback_args, **callback_kwargs)
Exception Type: TypeError at /api/practice-filter-subcategory-user
Exception Value: __init__() takes 1 positional argument but 2 were given更新urls.py :
from django.conf.urls import url
from yasn import views
urlpatterns = [
url(r'^api/practice-filter-user$', views.practiceFilter_User),
url(r'^api/practice-filter-subcategory-user$', views.practiceFilter_Subcategory_User),
]回答 1
Stack Overflow用户
回答已采纳
发布于 2021-09-05 16:39:52
是的，可以用许多参数进行滤波。Django Rest框架为我们提供了许多实现这一目标的选择。我会给你看其中一个：Filtering against query parameters。您可以重写基于类视图的api/practice-filter-subcategory-user?user=1&category=Math，以处理URL(如api/practice-filter-subcategory-user?user=1&category=Math)。
class practiceFilter_Subcategory_User(generics.ListAPIView):
serializer_class = PracticeSerializer
def get_queryset(self):
# If they are no parameters return all objects
queryset = Practice.objects.all()
# Retrieve the query parameters in the url
user_id = self.request.query_params.get('user_id')
category = self.request.query_params.get('category')
# If they are present
if user_id is not None and category is not None:
practice_filtered = exercises.filter(user__user_id=user_id)
# Filter by category, you can chain filter of course
practice_filtered = practice_filtered.filter(subcategory__category=category)
queryset = practice_filtered
return queryset这是DRF文件
非常重要:需要在url视图上调用.as_view()，如下所示：
from django.conf.urls import url
from yasn import views
urlpatterns = [
url(r'^api/practice-filter-user$', views.practiceFilter_User.as_view()),
url(r'^api/practice-filter-subcategory-user$', views.practiceFilter_Subcategory_User.as_view()),
]页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/69064918
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