Develop Reference

Django - Can I add a calculated field that only exists for a particular sub-set or occurences of my model?

Imagine that you have a model with some date-time fields that can be categorized depending on the date. You make an annotation for the model with different cases that assign a different 'status' depending on the calculation for the date-time fields:
#Models.py
class Status(models.TextChoices):
status_1 = 'status_1'
status_2 = 'status_2'
status_3 = 'status_3'
special_status = 'special_status'
class MyModel(models.Model):
important_date_1 = models.DateField(null=True)
important_date_2 = models.DateField(null=True)
calculated_status = models.CharField(max_length=32, choices=Status.choices, default=None, null=True, blank=False,)
objects = MyModelCustomManager()
And the manager with which to do the calculation as annotations:
# managers.py
class MyModelCustomManager(models.Manager):
def get_queryset(self):
queryset = super().get_queryset().annotate(**{
'status': Case(
When(**{'important_date_1' is foo, 'then':
Value(Status.status_1)}),
When(**{'important_date_2' is fii, 'then':
Value(Status.status_2)}),
When(**{'important_date_1' is foo AND 'importante_date_2' is whatever, 'then':
Value(Status.status_3)}),
# And so on and so on
)
}
)
return queryset
Now, here's where it gets tricky. Only one of these sub-sets of occurrences on the model requires an ADDITIONAL CALCULATED FIELD that literally only exists for it, that looks something like this:
special_calculated_field = F('important_date_1') - F('importante_date_2') #Only for special_status
So, basically I want to make a calculated field with the condition that the model instance must belong to this specific status. I don't want to make it an annotation, because other instances of the model would always have this value set to Null or empty if it were a field or annotation and I feel like it would be a waste of a row in the database.
Is there way, for example to do this kind of query:
>>> my_model_instance = MyModel.objects.filter(status='special_status')
>>> my_model_instance.special_calculated_field
Thanks a lot in advance if anyone can chime in with some help.

How to create multiple objects with different values at a time in django?

I need to create two models from a single template. Creating Product model is fine. The Product model has the ManyToOne relation with ProductVariant. But I got problem while creating ProductVariant model.
request.POST.getlist('names') this gives me the result like this ['name1','name2] and the same goes for all.
I want to create ProductVariant object with each values. How can I do this ? Also I think there is a problem while stroing a HStoreField. request.POST.getlist('attributes') gives the value like this ['a:b','x:z'] so I converted it into dictionary(but not sure it works).
UPDATE:
What I want is
attributes, names ... all will have the same number of items in the list.
For example if the name is ['a','b','c'] then weight will also have 3 values in the list [12,15,23] like this.
I want to create ProductVariant object 3 times since every list will have 3 items in the list. The first object will have field values from the list first item which is name=a,weight=12.. and for the second object values will be name=b, weight=15 like this.
How will it be possible? Or I should change the logic ? Any suggestions ?
models
class ProductVariant(models.Model):
name = models.CharField(max_length=255, blank=False, null=False)
product = models.ForeignKey(Product, on_delete=models.CASCADE)
attributes = HStoreField()
price = models.FloatField(blank=False, null=False, default=0.0)
views
product = product_form.save()
attributes = request.POST.getlist('attributes')
names = request.POST.getlist('name')
up = request.POST.getlist('price')
weight = request.POST.getlist('weight')
print(names, 'names')
# converting attributes into the dictionary for the HStore field
for attribute in attributes:
attributes_dict = {}
key, value = attribute.split(':')
attributes_dict[key] = value
ProductVariant.objects.create(name=name,...) # for each value I want to create this.

Answer for update:
names = ['a', 'b', 'c']
weights = [12, 15, 23]
params = zip(names, weights)
products = [ProductVariant(name=param[0], weight=param[1]) for param in params]
ProductVariant.objects.bulk_create(products)

I disagree with this approach, but if you really want to do it this way, ziping would be the way as #forkcs pointed out.
I would use Django to help me as much as possible, before i get there, please make this change. float != money
class ProductVariant(models.Model):
name = models.CharField(max_length=255, blank=False, null=False)
product = models.ForeignKey(Product, on_delete=models.CASCADE)
attributes = HStoreField()
price = models.DecimalField(blank=False, null=False, default=0, max_digits=6, decimal_places=2)
Once thats done, the form should look like this:
class ProductVariantForm(forms.ModelForm):
class Meta:
fields = ('name', 'product', 'attributes', 'price')
model = ProductVariant
ProductVariantFormSet = formset_factory(ProductVariantForm)
Note that I don't have to parse/clean/format attributes? Thats because Django did it for me ;)
And you can use it as follow IF you raname your fields and not use the same name multiple times: (instead of all your fields being called "attributes", you call them "form-X-attributes" where X is the number 0-infinity, example)
product = product_form.save()
formset = ProductVariantFormSet(data=request.POST)
if formset.is_valid():
instances = []
for form in formset:
if form.is_valid(): # this could probably be removed
instances.append(form.save())
For extra credit you can also do: (it shouldn't really matter)
product = product_form.save()
formset = ProductVariantFormSet(data=request.POST)
if formset.is_valid():
instances = []
for form in formset:
if form.is_valid(): # this could probably be removed
instances.append(form.save(save=False))
ProductVariant.objects.bulk_create(instances)
What do you gain? STANDARDS!!! AND compartmentalization! Everyone that knows Django knows what you did. All your clean logic will be placed in the right place (the form), and you'll be less error prone.
Ps. i wrote tests for you. https://gist.github.com/kingbuzzman/937a9d207bd937d1b2bb22249ae6bdb2#file-formset_example-py-L142
If you want more information on my approach, see the docs https://docs.djangoproject.com/en/3.1/topics/forms/formsets/

As for attributes, it could be reduced to one line like this:
attributes_dict = dict(map(lambda x: x.split(':'), attributes))
To create multiple objects you should either iterate and create one object at a time or use bulk_create:
for name in names:
ProductVariant.objects.create(name=name,...)
Or
ProductVariant.objects.bulk_create([ProductVariant(name=name) for name in names])

Best practice for this is using bulk_create method.
product_variants = [ProductVariant(name=name) for name in names]
ProductVariant.objects.bulk_create(product_variants)

django - prefetch only the newest record?

I am trying to prefetch only the latest record against the parent record.
my models are as such
class LinkTargets(models.Model):
device_circuit_subnet = models.ForeignKey(DeviceCircuitSubnets, verbose_name="Device", on_delete=models.PROTECT)
interface_index = models.CharField(max_length=100, verbose_name='Interface index (SNMP)', blank=True, null=True)
get_bgp = models.BooleanField(default=False, verbose_name="get BGP Data?")
dashboard = models.BooleanField(default=False, verbose_name="Display on monitoring dashboard?")
class LinkData(models.Model):
link_target = models.ForeignKey(LinkTargets, verbose_name="Link Target", on_delete=models.PROTECT)
interface_description = models.CharField(max_length=200, verbose_name='Interface Description', blank=True, null=True)
...
The below query fails with the error
AttributeError: 'LinkData' object has no attribute '_iterable_class'
Query:
link_data = LinkTargets.objects.filter(dashboard=True) \
.prefetch_related(
Prefetch(
'linkdata_set',
queryset=LinkData.objects.all().order_by('-id')[0]
)
)
I thought about getting LinkData instead and doing a select related but ive no idea how to get only 1 record for each link_target_id
link_data = LinkData.objects.filter(link_target__dashboard=True) \
.select_related('link_target')..?
EDIT:
using rtindru's solution, the pre fetched seems to be empty. there is 6 records in there currently, atest 1 record for each of the 3 LinkTargets
>>> link_data[0]
<LinkTargets: LinkTargets object>
>>> link_data[0].linkdata_set.all()
<QuerySet []>
>>>

The reason is that Prefetch expects a Django Queryset as the queryset parameter and you are giving an instance of an object.
Change your query as follows:
link_data = LinkTargets.objects.filter(dashboard=True) \
.prefetch_related(
Prefetch(
'linkdata_set',
queryset=LinkData.objects.filter(pk=LinkData.objects.latest('id').pk)
)
)
This does have the unfortunate effect of undoing the purpose of Prefetch to a large degree.
Update
This prefetches exactly one record globally; not the latest LinkData record per LinkTarget.
To prefetch the max LinkData for each LinkTarget you should start at LinkData: you can achieve this as follows:
LinkData.objects.filter(link_target__dashboard=True).values('link_target').annotate(max_id=Max('id'))
This will return a dictionary of {link_target: 12, max_id: 3223}
You can then use this to return the right set of objects; perhaps filter LinkData based on the values of max_id.
That will look something like this:
latest_link_data_pks = LinkData.objects.filter(link_target__dashboard=True).values('link_target').annotate(max_id=Max('id')).values_list('max_id', flat=True)
link_data = LinkTargets.objects.filter(dashboard=True) \
.prefetch_related(
Prefetch(
'linkdata_set',
queryset=LinkData.objects.filter(pk__in=latest_link_data_pks)
)
)

The following works on PostgreSQL. I understand it won't help OP, but it might be useful to somebody else.
from django.db.models import Count, Prefetch
from .models import LinkTargets, LinkData
link_data_qs = LinkData.objects.order_by(
'link_target__id',
'-id',
).distinct(
'link_target__id',
)
qs = LinkTargets.objects.prefetch_related(
Prefetch(
'linkdata_set',
queryset=link_data_qs,
)
).all()

LinkData.objects.all().order_by('-id')[0] is not a queryset, it is an model object, hence your error.
You could try LinkData.objects.all().order_by('-id')[0:1] which is indeed a QuerySet, but it's not going to work. Given how prefetch_related works, the queryset argument must return a queryset that contains all the LinkData records you need (this is then further filtered, and the items in it joined up with the LinkTarget objects). This queryset only contains one item, so that's no good. (And Django will complain "Cannot filter a query once a slice has been taken" and raise an exception, as it should).
Let's back up. Essentially you are asking an aggregation/annotation question - for each LinkTarget, you want to know the most recent LinkData object, or the 'max' of an 'id' column. The easiest way is to just annotate with the id, and then do a separate query to get all the objects.
So, it would look like this (I've checked with a similar model in my project, so it should work, but the code below may have some typos):
linktargets = (LinkTargets.objects
.filter(dashboard=True)
.annotate(most_recent_linkdata_id=Max('linkdata_set__id'))
# Now, if we need them, lets collect and get the actual objects
linkdata_ids = [t.most_recent_linkdata_id for t in linktargets]
linkdata_objects = LinkData.objects.filter(id__in=linkdata_ids)
# And we can decorate the LinkTarget objects as well if we want:
linkdata_d = {l.id: l for l in linkdata_objects}
for t in linktargets:
if t.most_recent_linkdata_id is not None:
t.most_recent_linkdata = linkdata_d[t.most_recent_linkdata_id]
I have deliberately not made this into a prefetch that masks linkdata_set, because the result is that you have objects that lie to you - the linkdata_set attribute is now missing results. Do you really want to be bitten by that somewhere down the line? Best to make a new attribute that has just the thing you want.

Tricky, but it seems to work:
class ForeignKeyAsOneToOneField(models.OneToOneField):
def __init__(self, to, on_delete, to_field=None, **kwargs):
super().__init__(to, on_delete, to_field=to_field, **kwargs)
self._unique = False
class LinkData(models.Model):
# link_target = models.ForeignKey(LinkTargets, verbose_name="Link Target", on_delete=models.PROTECT)
link_target = ForeignKeyAsOneToOneField(LinkTargets, verbose_name="Link Target", on_delete=models.PROTECT, related_name='linkdata_helper')
interface_description = models.CharField(max_length=200, verbose_name='Interface Description', blank=True, null=True)
link_data = LinkTargets.objects.filter(dashboard=True) \
.prefetch_related(
Prefetch(
'linkdata_helper',
queryset=LinkData.objects.all().order_by('-id'),
'linkdata'
)
)
# Now you can access linkdata:
link_data[0].linkdata
Ofcourse with this approach you can't use linkdata_helper to get related objects.

This is not a direct answer to you question, but solves the same problem. It is possible annotate newest object with a subquery, which I think is more clear. You also don't have to do stuff like Max("id") to limit the prefetch query.
It makes use of django.db.models.functions.JSONObject (added in Django 3.2) to combine multiple fields:
MainModel.objects.annotate(
last_object=RelatedModel.objects.filter(mainmodel=OuterRef("pk"))
.order_by("-date_created")
.values(
data=JSONObject(
id="id", body="body", date_created="date_created"
)
)[:1]
)

django in one queryset all and filter methods

I trying display in django app, in view last 5 item and also this items which has is_home set on True.
Please hint if this is 'nice' and correct way:
My model:
class Event(models.Model):
title = models.CharField(max_length=500)
date = models.DateField()
is_home = models.BooleanField(default=False)
My query in view:
context['event_list'] = Event.objects.filter(Q(Event.objects.all()) | Event.objects.filter(is_home=True))[:5]

context['event_list'] = Event.objects.filter(is_home=True).order_by(-id)[:5]

Simply use:
list(Event.objects.all().order_by('-id')[:5]) + list(Event.objects.filter(is_home=True))
Unfortunately, you cannot (as far as I can tell) combine queries after taking a slice, so conversion to lists is necessary.
If you really really want to have a QuerySet you can do:
Event.objects.filter(Q(id__in=Event.objects.all().order_by('-id')[:5].values_list('id', flat=True)) | Q(is_home=True))
Which is extremely ugly.

How to sort a Django QuerySet by (field, custom function, field)

I am looking for getting a QuerySet that is sorted by field1, function, field2.
The model:
class Task(models.Model):
issue_id = models.CharField(max_length=20, unique=True)
title = models.CharField(max_length=100)
priority_id = models.IntegerField(blank=True, null=True)
created_date = models.DateTimeField(auto_now_add=True)
def due_date(self):
...
return ageing
I'm looking for something like:
taskList = Task.objects.all().order_by('priority_id', ***duedate***, 'title')
Obviously, you can't sort a queryset by custom function. Any advise?

Since the actual sorting happens in the database, which does not speak Python, you cannot use a Python function for ordering. You will need to implement your due date logic in an SQL expression, as an Queryset.extra(select={...}) calculated field, something along the lines of:
due_date_expr = '(implementation of your logic in SQL)'
taskList = Task.objects.all().extra(select={'due_date': due_date_expr}).order_by('priority_id', 'due_date', 'title')
If your logic is too complicated, you might need to implement it as a stored procedure in your database.
Alternatively, if your data set is very small (say, tens to a few hundred records), you can fetch the entire result set in a list and sort it post-factum:
taskList = list(Task.objects.all())
taskList.sort(cmp=comparison_function) // or .sort(key=key_function)

The answer by #lanzz, even though seems correct, didn't work for me but this answer from another thread did the magic for me:
https://stackoverflow.com/a/37648265/6420686
from django.db.models import Case, When
ids = [list of ids]
preserved = Case(*[When(id=pk, then=pos) for pos, pk in enumerate(ids)])
filtered_users = User.objects \
.filter(id__in=ids) \
.order_by(preserved)

You can use sort in Python if the queryset is not too large:
ordered = sorted(Task.objects.all(), key=lambda o: (o.priority_id, o.due_date(), o.title))