Develop Reference

Unexpected keyword argument when working with several objects

I'm trying to work with severals objects to achieve an action.
My models.py
class LogBook(models.Model):
name = models.CharField()
class LogMessage(models.Model):
logbook = models.ForeignKey(LogBook, on_delete=models.CASCADE)
class LogDone(models.Model):
logmessage = models.ForeignKey(LogMessage)
done_status = models.BooleanField(default=False)
My view :
def logmessage_done(request, logmessage_id, log_id, token ):
log = get_object_or_404(LogBook, pk=log_id)
logmessages = LogMessage.objects.filter(logbook=log)
logdone = LogDone.objects.get_or_create(logmessage=logmessages)
logdone.done_status = True
logdone.update()
My url :
"done/<int:logmessage_id>/<int:log_id>/<str:token>"
What I want to achieve :
I want to change the status of the logdone object which is link to the logmessage object but I am not sure I have access object correctly.
What error I have :
The QuerySet value for an exact lookup must be limited to one result using slicing.

Change your view like this:
def logmessage_done(request, logmessage_id, log_id, token ):
log = get_object_or_404(LogBook, pk=log_id)
logmessages = LogMessage.objects.filter(logbook=log)
for log_message in logmessages:
LogDone.objects.update_or_create(logmessage=log_message,defaults={"done_status": True})
Here , log returns a single object with id . logmessages returns a queryset with logbook = the log returned in first query. Have to use update_or_create method

Django - cannot retrieve just one record in multi-part filter on model with multiple relations

I can't seem to isolate a single record from this query:
subcust = OwnerCustom.objects.get(carcustom=ncset, owner=sset)
This is the error:
OwnerCustom matching query does not exist
In the actual data, there is only actually one matching record in OwnerCustom for each record in CarCustom. It's supposed to be a kind of many-to-many where there are standard differences listed in CarCustom for each Car, and each owner may maintain their own customizations (overrides) or those default OwnerCustom entries.
Note, there are many different Owner of the same Car. And of course, I'm not actually doing cars, this is a renaming from the original purpose.
Here's the relevant models:
class Car(models.Model):
car_name = models.CharField(max_length=50)
class CarCustom(models.Model):
car = models.ForeignKey(Car, models.PROTECT)
class Owner(models.Model):
car = models.ForeignKey(Car, models.PROTECT)
class OwnerCustom(models.Model):
owner = models.ForeignKey(Owner, models.PROTECT)
carcustom = models.ForeignKey(CarCustom, models.PROTECT)
name = models.CharField(max_length=50)
And the code:
car_queryset = Car.objects.filter(car_name="fancy car")
for nset in car_queryset:
owner_queryset = Owner.objects.filter(car=nset)
for sset in owner_queryset :
carcustom_queryset = CarCustom.objects.filter(car=nset)
for ncset in carcustom_queryset:
subcust = OwnerCustom.objects.get(carcustom=ncset, owner=sset)
I've tried stuff like:
subcust = OwnerCustom.objects.filter(carcustom=ncset, owner=sset).first()
Which gives me a NoneType, and then tried:
subcust = OwnerCustom.objects.filter(carcustom=ncset, owner=sset)[:1].get()
Which gives "matching query does not exist" and this:
subcust = OwnerCustom.objects.filter(carcustom=ncset, owner=sset)[0]
Gives "list index out of range"
UPDATE: I CAN get a working function by using code like this, but I would think since there is only one (guaranteed by application) matching record possible for OwnerCustom.objects.filter(carcustom=ncset, owner=sset) that I could find a better way to fetch it:
car_queryset = Car.objects.filter(car_name="fancy car")
for nset in car_queryset:
owner_queryset = Owner.objects.filter(car=nset)
for sset in owner_queryset :
carcustom_queryset = CarCustom.objects.filter(car=nset)
for ncset in carcustom_queryset:
subcust_queryset = OwnerCustom.objects.filter(carcustom=ncset, owner=sset)
for subcust in subcust_queryset :
logger.info(subcust.name)

I need to query for a set of objects whose primary keys are contained inside of a list

As the title says, I need a way to perform this query. I have tried the following:
user_list_ids = []
user_lists = []
user_entries = OwnerEntry.objects.filter(name=request.user)
for user in user_entries:
user_list_ids.append(user.list_id)
user_lists = ListEntry.objects.filter(id__in=user_list_ids)
for user in user_entries:
user_list_ids.append(user.list_id)
user_lists = ListEntry.objects.filter(id__in=user_list_ids)
However, I get an error on the last line: int() argument must be a string or a number, not 'ListEntry'
Here are the relevant models:
class OwnerEntry(models.Model):
name = models.CharField(max_length=32)
list_id = models.ForeignKey(ListEntry)
class Meta:
ordering = ('name',)
class ListEntry(models.Model):
name = models.CharField(max_length=64)
# active_date = models.DateTimeField('date of last list activity')
expire_date = models.DateField('date of expiration')
create_date = models.DateField('date created')

to answer your question directly, please note that you have a list_id rather than list as a ForeignKey name (OwnerEntry model). In order to actually extract the fk value, you should use list_id_id instead (or rename list_id to list ;))
Please also note that django supports object references, like so:
someowner = OwnerEntry.objects.get( ... )
ownerslist = someowner.listentry_set.all()
cheers!

You can define OwnerEntry's foreign key to ListEntry as :
list_id = models.ForeignKey(ListEntry, related_query_name='owner_entry')
and then do this one-liner in your code:
user_lists = ListEntry.objects.filter(owner_entry__name=request.user)
What this does is exactly filter every ListEntry which has at least one owner_entry whose name is equal to request.user's.
The redefinition of the foreign key is just for the sake of giving a nice name to the query attribute.
For more details on queries that work with backward relationships: https://docs.djangoproject.com/en/dev/topics/db/queries/#lookups-that-span-relationships

Django ManyToManyField Error when saving in admin?

What is wrong with my code?
class Group(ImageModel):
title = models.CharField(verbose_name = "Title", max_length=7)
photos = models.ManyToManyField('Photo', related_name='+',
verbose_name=_('Photo'),
null=True, blank=True)
.....
pid = Photo.objects.get(image = str_path)
gid= Group.objects.get(id = self.id)
self.save_photos(gid, pid)
....
def save_photos(self, gid, pid):
group_photo = GroupPhotos(groupupload=gid.id,
photo=pid.id
)
group_photo.save()
and my GroupPhotos models is:
class GroupPhotos(models.Model):
groupupload = models.ForeignKey('Group')
photo = models.ForeignKey('Photo')
class Meta:
db_table = u'group_photos'
when i want to save it from admin panel i am getting value error sth like this:
Cannot assign "38": "GroupPhotos.groupupload" must be a "Group" instance.
with group_photo = GroupPhotos(groupupload=gid, photo=pid) defination it is working but there is no any changes in GroupPhotos table(group_photos). printing this print pid.id,' >>> ',gid.id i am getting true relation...
UPDATE:
I have been working since morning, but no progress... i have also tried this but nothing changed:
pid = Photo.objects.get(image = str_path)
ger = Group.objects.get(id = self.id)
ger.title = self.title
ger.save()
ger.photos.add(pid)

The error is here:
group_photo = GroupPhotos(groupupload=gid.id, photo=pid.id)
The arguments to groupupload and photo should be instances of Group and Photo respectively. Try the following:
group_photo = GroupPhotos(groupupload=gid, photo=pid)
In other words, when creating an object you need to pass arguments of the expected type and not an integer (which may be the primary key key of the desired object but it also might not, which is why you need to pass an object of the correct type).

i have solved my problem with adding through option to my manytomanyfield:
photos = models.ManyToManyField('Photo', related_name='+',
verbose_name=_('Photo'),
null=True, blank=True, through=GroupPhotos)
some info about ManyToManyField.through here:
Django will automatically generate a table to manage many-to-many
relationships. However, if you want to manually specify the
intermediary table, you can use the through option to specify the
Django model that represents the intermediate table that you want to
use.
The most common use for this option is when you want to associate extra data with a many-to-many relationship.

(Django) Cannot assign "u'1'": "StaffProfile.user" must be a "User" instance

I have a model like below:
class StaffProfile(models.Model):
user = models.ForeignKey(User)
maas = models.FloatField()
maas_gunu = models.CharField(max_length=5)
When I try to insert data with a code like below:
staffprofilesay = StaffProfile.objects.filter(user = user_id).count()
if staffprofilesay > 0:
staffprofile = StaffProfile.objects.get(user = user_id)
else:
staffprofile = StaffProfile()
staffprofile.user = user_id
staffprofile.maas = calisan_formu.cleaned_data["maas"]
staffprofile.maas_gunu = calisan_formu.cleaned_data["maas_gunu"]
staffprofile.save()
I get an error like this:
Cannot assign "u'1'": "StaffProfile.user" must be a "User" instance.
What am I supposed to do?
PS: I'm using Django's User model

You need to assign a User object e.g.
from django.contrib.auth.models import User
user = User.objects.get(id=user_id)
staffprofile.user = user

user needs to be an instance of the User model, not a unicode object (which is what you are passing it).

Yes you have to pass User instance in staffprofile.user = user_id user id place.
As #david-s pointed out in a comment, if you don't have a user instance, you have to fetch from DB with an additional query.
Instead you can directly do is
staffprofile.user_id = user_id because Django behind the scene append _id in table for foreign keys so staffprofile.user will end staffprofile.user_id