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django import_export | data not exported from database

I am using Django-import_export for exporting data. I have used the code given below that's not working correctly. It exported only dehydrated data instead of given fields.
class MemberResource(resources.ModelResource):
Brand=Field()
class meta:
model = model
fields=('title','Brand')
def dehydrate_Brand(self, obj):
return str(obj.Brand.title)
class modelAdmin(ImportExportModelAdmin):
resource_class = MemberResource
list_display=['title','Model_code','Chipset','chipset_description','Brand','categories']
search_fields = ['title','Model_code','Chipset',]
fields=('title','Model_code','Chipset','chipset_description','image','Brand','Cat')
admin.site.register(model,modelAdmin)

The name of the Meta subclass is Meta, not meta, so the ModelResource should look like:
class MemberResource(resources.ModelResource):
Brand=Field()
class Meta:
model = Member
fields = ('title','Brand')
def dehydrate_Brand(self, obj):
return str(obj.Brand.title)

Filter field of manytomany field by a list

I would like to do the following with Django REST Framework:
Filter results based on a field of a manytomany field.
The query would look like this:
https://endpoint.com/api/artwork/?having_style=Modern,Contemporary
I would expect the result to contain all ArtWork objects which contain a relation to a Style object with name "Modern", "Contemporary" or both.
The code below is not working and I don't know why.
models.py
class Style(models.Model):
name = models.CharField(validators=[validate_style], max_length=100, unique=True)
class ArtWork(models.Model):
styles = models.ManyToManyField(Style, default=None)
filters.py
class ArtWorkFilter(filters_rest.FilterSet):
having_style = django_filters.Filter(field_name="styles__name", lookup_expr='in')
class Meta:
model = ArtWork
fields = ['having_style']
class StyleSerializer(serializers.ModelSerializer):
class Meta:
model = Style
fields = ('name',)
class ArtWorkSerializer(serializers.ModelSerializer):
styles = StyleSerializer(many=True)
class Meta:
model = ArtWork
fields = ('styles'/)
views.py
class ArtWorkViewSet(viewsets.ModelViewSet):
permission_classes = []
queryset = ArtWork.objects.all()
serializer_class = ArtWorkSerializer
filter_backends = [filters_rest.DjangoFilterBackend,]
filterset_class= ArtWorkFilter
pagination_class = CursorSetPagination
Thank you in advance!
Solution
I solved it by changing the ArtWorkFilter to
filters.py
class ArtWorkFilter(filters_rest.FilterSet):
having_style = django_filters.Filter(field_name="styles__name", lookup_expr='in')
class Meta:
model = ArtWork
fields = ['having_style']
def filter_by_style_name(self, queryset, name, value):
list_styles = value.split(',')
return queryset.filter(styles__name__in=list_styles)

Try adding method param in Filter declaration. Something like:
class ArtWorkFilter(filters_rest.FilterSet):
having_style = django_filters.Filter(field_name="styles__name", lookup_expr='in')
class Meta:
model = ArtWork
fields = ['having_style']
def filter_by_style_name(self, queryset, name, value):
list_styles = value.split(',')
return queryset.filter(styles__name__in=list_styles)

CartItem.objects.filter(cart=cart, product=product, attribute__in=attribute_list).annotate(num_attr=Count('attribute')).filter(num_attr=len(attribute_list))

Parametrize a Python class

Is there any way to parametrize a class in Python? The parametrized class may look something like this:
class FIELDSerializer:
FIELD = serializers.CharField(source='get_FIELD_display', required=False)
class Meta:
model = None
fields = {FIELD}
Which would need to create the following three classes:
class NameSerializer:
name = serializers.CharField(source='get_name_display', required=False)
class Meta:
model = None
fields = {'name'}
class CategorySerializer:
category = serializers.CharField(source='get_category_display', required=False)
class Meta:
model = None
fields = {'category'}
class StateSerializer:
state = serializers.CharField(source='get_state_display', required=False)
class Meta:
model = None
fields = {'state'}
Is this possible or not?

You can do what you want with a factory function, although it's not completely trivial to get the internal variables (attributes) as you want them:
def factory(FIELDname):
class FIELDSerializer:
class Meta:
model = None
fields = {FIELDname}
settatr(FIELDSerializer, FIELDname, serializers.CharField(source=f'get_{FIELDname}_display', required=False))
return FIELDSerializer
CategorySerializer = factory('category')
StateSerializer = factory('state')
NameSerializer = factory('name')
The setattr allows us to set the name of attribute to the FIELDname string. (Thanks to #Code-Apprentice and #juanpa.arrivillaga for this idea.)
I don't know if there's any easy way to avoid the repetition of the field name and the desired class name when you call the factory without using something like exec (which is perfectly legal but usually leaves programmers with a bad taste in their mouths).

django rest framework - how to add a static field value for a serializer

I need to add a static field to my serializer. It should always return the same value, regardless of the passed object. Currently I implemented it like so:
class QuestionSerializer(serializers.ModelSerializer):
type = serializers.SerializerMethodField()
#staticmethod
def get_type(obj):
return 'question'
class Meta:
model = Question
fields = ('type',)
But is there a simpler way to do it, without the SerializerMethodField?

using a ReadOnlyField worked for me:
class QuestionSerializer(serializers.ModelSerializer):
type = serializers.ReadOnlyField(default='question')
class Meta:
model = Question
fields = ('type',)
https://www.django-rest-framework.org/api-guide/fields/#readonlyfield

The only alternative would be to override to_representation and add the value there:
def to_representation(self, obj):
data = super().to_representation(obj)
data['type'] = 'question'
return data
Not a much better option though.

You can use serializers.HiddenField
class QuestionSerializer(serializers.ModelSerializer):
type = serializers.HiddenField(default='question')
class Meta:
model = Question
fields = ('type',)
http://www.django-rest-framework.org/api-guide/fields/#hiddenfield

If you don't care about adding a line to your model, it would be easier to add static field to your model.
class Question(models.Model):
type = 'question'
class QuestionSerializer(serializers.ModelSerializer):
class Meta:
model = Question
fields = ('type',)

Inherit and modify a `Meta` class

So I have a base ItemTable, and then a number of Tables that inherit from it. I don't seem to be able to modify the Meta class. I tried just including the meta class normally and it didn't work, then I found this bug report and implemented it below. It fails silently: the tables render only with the columns from the parent meta class.
class ItemTable(tables.Table):
class Meta:
model = Item
attrs = {"class":"paleblue"}
fields = ('name', 'primary_tech', 'primary_biz', 'backup_tech', 'backup_biz')
class ApplicationTable(ItemTable):
def __init__(self, *args, **kwargs):
super(ApplicationTable, self).__init__(*args, **kwargs)
class Meta(ItemTable.Meta):
model = Application
fields += ('jira_bucket_name',)
EDIT: Code amended as shown. I now get a NameError that fields is not defined.

Try:
class ApplicationTable(ItemTable):
class Meta:
model = Application
fields = ItemTable.Meta.fields + ('jira_bucket_name',)
You'll have the same problems extending Meta in a table, as you will in a normal Django model.

You didnt add , (comma) to one-element tuple. Try to change this line Meta.attrs['fields'] += ('jira_bucket_name') in ApplicationTable to:
Meta.attrs['fields'] += ('jira_bucket_name',)
if it didnt help try to create Meta class outsite model class definition:
class ItemTableMeta:
model = Item
attrs = {"class":"paleblue"}
fields = ('name', 'primary_tech', 'primary_biz', 'backup_tech', 'backup_biz')
class ApplicationTableMeta(ItemTableMeta):
model = Application
fields = ItemTableMeta.fields + ('jira_bucket_name',)
class ItemTable(tables.Table):
#...
Meta = ItemTableMeta
class ApplicationTable(ItemTable):
#...
Meta = ApplicationTableMeta

You may need to take this up with the django-tables author. This is not a problem with standard Django.