I've been using the built-in Django comments system which has been working great. On a particular page I need to list the latest X comments which I've just been fetching with:
latest_comments =
Comment.objects.filter(is_public=True, is_removed=False)
.order_by('submit_date').reverse()[:5]
However I've now introduced a Boolean field 'published' into the parent object of the comments, and I want to include that in the query above. I've tried using the content_type and object_pk fields but I'm not really getting anywhere. Normally you'd do something like:
Comment.objects.filter(blogPost__published=True)
But as it is not stored like that I am not sure how to proceed.
posts_ids = BlogPost.objects.filter(is_published=True).values_list('id', flat=True) #return [3,4,5,...]
ctype = ContentType.objects.get_for_model(BlogPost)
latest_comments = Comment.objects.filter(is_public=True, is_removed=False, content_type=ctype, content_object__in=posts_ids).order_by('-submit_date')[:5]
Comments use GenericForeignKey to store the relation to parent object. Because of the way generic relations work related lookups using __<field> syntax are not supported.
You can accomplish the desired behaviour using the 'in' lookup, however it'll require lot of comparisons when there'll be a lot of BlogPosts.
ids = BlogPost.objects.filter(published=True).values_list('id', flat=True) # Get list of ids, you would probably want to limit number of items returned here
content_type = ContentType.objects.get_for_model(BlogPost) # Becasue we filter only comments for BlogPost
latest_comments = Comment.objects.filter(content_type=content_type, object_pk__in=ids, is_public=True, is_removed=False, ).order_by('submit_date').reverse()[:5]
See the Comment model doc for the description of all fields.
You just cannot do that in one query. Comments use GenericForeignKey. Documentation says:
Due to the way GenericForeignKey is implemented, you cannot use such
fields directly with filters (filter() and exclude(), for example) via
the database API.
Related
I am trying to generate nested HATEOAS links in a serializer using the drf-nested-routes package. My current setup would be as follows:
/resource_a/<pk>
/resource_a/<pk>/resource_b/<pk>
/resource_a/<pk>/resource_b/<pk>
/resource_a/<pk>/resource_b/<pk>/resource_c
I am unable to create a HyperlinkedIdentityField that points to the last route. According to the documentation, one can create hyperlinked fields like this:
nameservers = HyperlinkedIdentityField(
view_name='domain-nameservers-list',
lookup_url_kwarg='domain_pk'
)
Or
nameservers = NestedHyperlinkedRelatedField(
many=True,
read_only=True, # Or add a queryset
view_name='domain-nameservers-detail'
parent_lookup_url_kwargs={'domain_pk': 'domain__pk'}
)
But these approaches fail when trying to reach a resource that is 2 layers deep in the URL hierarchy. The first method is not compatible, as it does not allow to add a second lookup_url_kwarg, and as for the second one, it throws an exception (ImproperlyConfigured) when configuring with the (in my opinion) proper attributes (resource_a__pk, resource_b__pk).
Is this at all possible with this package? Otherwise I will resort to a simpler solution using a SerializerMethodField:
resource_c = serializers.SerializerMethodField()
def get_resource_c(self, obj):
url = reverse('resource_b-resource_c-list', kwargs=dict(resource_a_pk=obj.resource_a.pk, resource_b_pk=obj.pk))
return self.context['request'].build_absolute_uri(url)
Thanks in advance!
I've done this before using NestedHyperlinkedRelatedField and it definitely works. My guess is that your configuration is not correct. One thing I noticed is that you use parent_lookup_url_kwargs while in my case I use parent_lookup_kwargs.
Based on your explanation I think it needs to look like this
NestedHyperlinkedRelatedField(...,
parent_lookup_kwargs={
'resource_a_pk': '<how to reach resource_a pk from resource_b>'})
I'm using sqlalchemy and have two models, Article and Tag, it's a many-to-many relation.
When I add articles using Flask-Admin, I want just part of tags available (related on user permission) instead of all tags.
any idea? Thanks
Probably the best way to do this is to use dynamic relationship loaders. Simply use lazy='dynamic' in your relationship definition:
posts = relationship(Post, lazy="dynamic")
This returns you a query object instead of a collection of objects, so you can then query it directly:
posts = jack.posts.filter(Post.headline=='this is a post')
You could also achieve what you want with discriminator columns or something, but that is likely overkill.
sounds like you need ModelView.get_query:
class MyView(ModelView):
def get_query(self,*args,**kwargs):
return super(MyView,self).get_query(*args,**kwargs).filter_by(current_user.can_view=True)
I have a lot of objects to save in database, and so I want to create Model instances with that.
With django, I can create all the models instances, with MyModel(data), and then I want to save them all.
Currently, I have something like that:
for item in items:
object = MyModel(name=item.name)
object.save()
I'm wondering if I can save a list of objects directly, eg:
objects = []
for item in items:
objects.append(MyModel(name=item.name))
objects.save_all()
How to save all the objects in one transaction?
as of the django development, there exists bulk_create as an object manager method which takes as input an array of objects created using the class constructor. check out django docs
Use bulk_create() method. It's standard in Django now.
Example:
Entry.objects.bulk_create([
Entry(headline="Django 1.0 Released"),
Entry(headline="Django 1.1 Announced"),
Entry(headline="Breaking: Django is awesome")
])
worked for me to use manual transaction handling for the loop(postgres 9.1):
from django.db import transaction
with transaction.atomic():
for item in items:
MyModel.objects.create(name=item.name)
in fact it's not the same, as 'native' database bulk insert, but it allows you to avoid/descrease transport/orms operations/sql query analyse costs
name = request.data.get('name')
period = request.data.get('period')
email = request.data.get('email')
prefix = request.data.get('prefix')
bulk_number = int(request.data.get('bulk_number'))
bulk_list = list()
for _ in range(bulk_number):
code = code_prefix + uuid.uuid4().hex.upper()
bulk_list.append(
DjangoModel(name=name, code=code, period=period, user=email))
bulk_msj = DjangoModel.objects.bulk_create(bulk_list)
Here is how to bulk-create entities from column-separated file, leaving aside all unquoting and un-escaping routines:
SomeModel(Model):
#classmethod
def from_file(model, file_obj, headers, delimiter):
model.objects.bulk_create([
model(**dict(zip(headers, line.split(delimiter))))
for line in file_obj],
batch_size=None)
Using create will cause one query per new item. If you want to reduce the number of INSERT queries, you'll need to use something else.
I've had some success using the Bulk Insert snippet, even though the snippet is quite old.
Perhaps there are some changes required to get it working again.
http://djangosnippets.org/snippets/446/
Check out this blog post on the bulkops module.
On my django 1.3 app, I have experienced significant speedup.
bulk_create() method is one of the ways to insert multiple records in the database table. How the bulk_create()
**
Event.objects.bulk_create([
Event(event_name="Event WF -001",event_type = "sensor_value"),
Entry(event_name="Event WT -002", event_type = "geozone"),
Entry(event_name="Event WD -001", event_type = "outage") ])
**
for a single line implementation, you can use a lambda expression in a map
map(lambda x:MyModel.objects.get_or_create(name=x), items)
Here, lambda matches each item in items list to x and create a Database record if necessary.
Lambda Documentation
The easiest way is to use the create Manager method, which creates and saves the object in a single step.
for item in items:
MyModel.objects.create(name=item.name)
I want to use django's admin filter on the list page.
The models I have are something like this:
class Location(model):
name = CharField()
class Inquiry(Model):
name = CharFiled()
location = ManyToManyField(Location)
Now I want to filter Inquiries, to display only those that contain relation to specific Location object. If I use
class InqAdmin(ModelAdmin):
list_filter = ['location', ]
admin.site.register(Inquiry, InqAdmin)
the admin page displays me the list of all Locations and allows to filter.
What I would like to get, is to get list of only those locations that have some Inquiries in relation to them (so I don't ever get the empty list result after filtering).
How can this be done?
You could create a custom manager for Locations that only returns Locations that have an Inquiry associated with them. If you make this the default manager, the admin will use it.
Only caveat is that you'll need create another manager that returns all Locations and use that in the rest of your app whenever you want to retrieve Locations that don't have an associated Inquiry.
The managers section in the Django docs is quite good, and should be all you need to get this set up.
EDIT:
sienf brings up a good point. Another way to accomplish this would be to define a subclass of django.contrib.admin.SimpleListFilter, and write the queryset method to filter out Inquiries with empty Locations. See https://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_filter
I'd like to have access to one my model field verbose_name.
I can get it by the field indice like this
model._meta._fields()[2].verbose_name
but I need to get it dynamically. Ideally it would be something like this
model._meta._fields()['location_x'].verbose_name
I've looked at a few things but I just can't find it.
For Django < 1.10:
model._meta.get_field_by_name('location_x')[0].verbose_name
model._meta.get_field('location_x').verbose_name
For Django 1.11 and 2.0:
MyModel._meta.get_field('my_field_name').verbose_name
More info in the Django doc
The selected answer gives a proxy object which might look as below.
<django.utils.functional.__proxy__ object at 0x{SomeMemoryLocation}>
If anyone is seeing the same, you can find the string for the verbose name in the title() member function of the proxy object.
model._meta.get_field_by_name(header)[0].verbose_name.title()
A better way to write this would be:
model._meta.get_field(header).verbose_name.title()
where header will be the name of the field you are interested in. i.e., 'location-x' in OPs context.
NOTE: Developers of Django also feel that using get_field is better and thus have depreciated get_field_by_name in Django 1.10. Thus I would suggest using get_field no matter what version of Django you use.
model._meta.get_field_by_name('location_x')[0].verbose_name
You can also use:
Model.location_x.field.verbose_name
Model being the class name. I tested this on my Animal model:
Animal.sale_price.field.verbose_name
Animal.sale_price returns a DeferredAttribute, which has several meta data, like the verbose_name
Note: I'm using Django 3.1.5
If you want to iterate on all the fields you need to get the field:
for f in BotUser._meta.get_fields():
if hasattr(f, 'verbose_name'):
print(f.verbose_name)
# select fields for bulk_update : exclude primary key and relational
fieldsfields_to_update = []
for field_to_update in Model._meta.get_fields():
if not field_to_update.many_to_many and not field_to_update.many_to_one and not field_to_update.one_to_many and not field_to_update.one_to_one and not field_to_update.primary_key and not field_to_update.is_relation :
fields_to_update = fields_to_update + [field_to_update.name]
Model.objects.bulk_update(models_to_update , fields_to_update)