Indexing not supported? - python

I get this error message:
TypeError: 'City' object does not support indexing
when this is my model:
class City(db.Model):
region = db.ReferenceProperty()
name = db.StringProperty()
image = db.BlobProperty()
vieworder = db.IntegerProperty()
areacode = db.IntegerProperty()
and this is my query
items = Item.all().filter('modified >', timeline).filter('published =', True).order('-modified').filter('cities =',city[0].key()).fetch(PAGESIZE + 1)`
Can you tell my how I should make my query or model my class definitions? I use listproperty(db.Key) to model relationships with references and I thought I could filter like the above query since it worked for one item. What should I do?

city[0].key()
is the cause of your error. Chances are you thought city was a list, but it's actually a single item, or some such.

Related

Django ORM: Object is not iterable (Error)

Can anyone explain why this is iterable:
User.objects.all()
this is valid and gives me a value (The current user's alias. session is storing the user id):
User.objects.get(id = request.session['currentuser']).alias)
But this is giving me the error saying it is 'not iterable?':
Poke.objects.get(user = User.objects.get(id = request.session['currentuser']).alias)
(This code is supposed to get a list of Poke entries where the user column matches the current user's alias.)
Here is the Poke model. It does not use ForeignKeys, as I was having trouble setting two of them without errors.
class Poke(models.Model):
id = models.IntegerField(primary_key=True)
user = models.CharField(max_length=100)
poker = models.CharField(max_length=100)
pokes = models.IntegerField()
class Meta:
app_label = "poke_app"
Get will retrieve a single object and therefore the result will not be iterable. See documentation.
Do you see an integer value when you print(request.session['currentuser'])?
If you will see a string then you shoud give an integer value
EX: userobj = User.objects.get(id=uid)
Oh sory
User.objects.get(id = request.session['currentuser']).alias)
You open ( and closed it after ['currentuser']) but why you close ) again after .alias ?

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)

Django - using Foreign key mode attribute

I have two Django models and connected via Foreignkey element and in the second model I need to use the firs model's attribute - example (pseudocode):
Class Category(models.Model):
c_attribute = "Blue"
Class Object(models.Model):
o_category = models.ForeignKey(Category)
o_color = o_category.c_attribute
The key here is the last line - I got error saying that ForeignKey object has no attribute c_attribute.
Thanks
Because o_category is a Key to a Category, not a Category itself!
you can check by type(o_category) to check is not Category!
so you have access to related Cateogry of a Object in other parts of application when connected to database.for example in shell you can write:
c = Category()
c.save()
o = Object(o_category = c, ...) #create Object with desired params
... #some changes to o
o.save()
o.o_category.c_attribute #this will work! :)
You can use to_field='', but that might give you an error as well.
Class Object(models.Model):
o_category = models.ForeignKey(Category)
o_color = models.ForeignKey(Category, to_field="c_attribute")
The best thing is do create a function in your Object model, that would get you the categories c_attribute like so:
def get_c_attribute(self):
return self.o_category.c_attribute

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.

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