Django values_list with choices field - django

The values_list in filtering object, really helps me a lot in providing solution within django view.
My code is like the following and this one works:
values_list_ac = realdata.objects.filter(product = '1').values_list('company', 'brand', 'created_by__username')
while username is the field exists in different model outside the current realdata model.
But the following code doesn't work, for I want to show the value of ac_type, which based on choices field within the same realdata model. (I try to solve it by using the same solution which work in template):
values_list_ac = realdata.objects.filter(product = '1').values_list('company', 'brand', 'created_by__username', 'get_ac_type_display')
Is there a solution other than get_ac_type_display to show the field value?
I really appreciate for some shed of light.
Edit:
This my model:
class realdata(models.Model):
company = models.CharField(max_length=60, verbose_name="Company")
brand = models.CharField(_('brand'), max_length=60)
model = models.CharField(max_length=60)
type_choices = (
(u'1', u'Inverter'),
(u'2', u'Non-Inverter'),
)
ac_type = models.CharField(max_length=60, verbose_name="Type", choices=type_choices)
created_by = models.ForeignKey(User)
Many Thanks!

The values_list function will just get the values stored in the database. When defining choices on your model's field, it will store the first value of the tuple, hence this will be what you'll retrieve.
This means that you have to look at the choices tuple to determine the display value for the item. The purpose of the get_foo_display is to give you this human-readable value, but it needs a model instance to work on.
So, a solution to resolving this yourself would be to inspect the choices, and convert the data accordingly. The following should be able to do this:
result = []
for p in realdata.objects.filter(product='1').values_list(
'company', 'brand', 'created_by__username', 'ac_type'):
choice = {k: v for k, v in realdata.type_choices}[p[-1]]
result.append(list(p[:-1]) + [choice])
The result variable will contain the converted list. The new variable is needed because the values_list function will return a list of tuples; the latter being unmutable. Also, take care to have the value you'll want to resolve as the last item in your values_list call, or adapt the above to match.

Related

in Django: How to get and display in template verbose_name of a model that their value is true, while name of field unknown?

I have a model than has a lot of models.BooleanField declarations.
class a_lot_of_booleans(model.Models):
old_or_new = models.BooleanField(default=False,verbose_name="is it an old or a new item")
product_for_teens = models.BooleanField(default=False,verbose_name="is this product for teens")
in_original_package = models.BooleanField(default=False,verbose_name="is this product in original package?")
Which then is used in some other classes like:
class product_for_sale(a_lot_of_booleans):
U_Id = models.OneToOneField(User, on_delete=models.CASCADE, primary_key=True)
product_name=models.CharField(max_length=50)
class product_for_buying(a_lot_of_booleans):
U_Id = models.OneToOneField(User, on_delete=models.CASCADE, primary_key=True)
product_name=models.CharField(max_length=50)
The class a_lot_of_booleans might change over time. Some booleans might get added some might get removed. What I need is to display a list of several entries, of the verbose_name, of only the true fields on one of the classes that inherit the a_lot_of_booleans class and the value of product_name, belonging to specific user.
What I"m trying in the views.py is the following:
def view_rfps(request):
list=product_for_sale.objects.all().filter(U_Id=request.user)
for item in list:
values=item._meta.fields
for value in values:
res=item.objects.filter(**{value:'True'}) ##<< lines that fail
print(res)
the above code fails on res=item.objects.filter(**{value:'True'}) on "Manager isn't accessible via search_for_constructor_rfp instances"
The idea later to pass on the res variable to view, however I cannot pass this point.
I have several items in list and for every list several boolean fields, that I"m not sure what they names gonna be in a future, so I cannot just use product_for_sale. in template later.
Any suggestion how to display the verbose name of unknown boolean field name ?
edit
Found a way:
def view_rfps(request):
list=product_for_sale.objects.all().filter(U_Id=request.user)
for item in list:
values=item._meta.fields
for value in values:
temp=getattr(item,value.name)
if temp:
print(value.verbose_name)
But if someone knows more efficient way, I"d love to hear
This is what worked for me:
def view_rfps(request):
list=product_for_sale.objects.all().filter(U_Id=request.user)
for item in list:
values=item._meta.fields
for value in values:
temp=getattr(item,value.name)
if temp:
print(value.verbose_name)

Return a value of a model based on the max value of another field of the model

I have a model in Django with two fields: name and meter. I have created a views file, and I want to find and print the name with the biggest meter. The code below find and prints the biggest meter itself which is good, but not what I want. I want to print the name that is related to that meter. Any idea how to do so?
My Views:
def details (request):
top_meter = Energy_Consuption.objects.aggregate(Max('meter'))
return render(request,'details.html', {'top_meter':top_meter})
(The view is connected to an HTML page.)
My Model:
class Energy_Consuption(models.Model):
name=models.IntegerField()
meter=models.IntegerField()
class Meta:
db_table:'Energy_Consuption'
You can do something like this:
Energy_Consuption.objects.filter(
meter=Energy_Consuption.objects.aggregate(Max('meter'))['meter__max']
).values_list('name', flat=True)
This will return the list of the names of the Energy_Consuption objects that matches the max value. Note that you need filter here, because there can be more than one Energy_Consuption with the max value.
You can just sort the QuerySet by meter and take the latest item.
Additionally, you can use getattr in case of QuerySet being empty.
biggest_item = EnergyConsumption.objects.latest('meter')
biggest_name = getattr(biggest_item, 'name', None)
If your meter field is not unique, you can provide additional fields to the latest method to disambiguate:
biggest_item = EnergyConsumption.objects.latest('meter', 'id')

Query intermediate through fields in django

I have a simple Relation model, where a user can follow a tag just like stackoverflow.
class Relation(models.Model):
user = AutoOneToOneField(User)
follows_tag = models.ManyToManyField(Tag, blank=True, null=True, through='TagRelation')
class TagRelation(models.Model):
user = models.ForeignKey(Relation, on_delete=models.CASCADE)
following_tag = models.ForeignKey(Tag, on_delete=models.CASCADE)
pub_date = models.DateTimeField(default=timezone.now)
class Meta:
unique_together = ['user', 'following_tag']
Now, to get the results of all the tags a user is following:
kakar = CustomUser.objects.get(email="kakar#gmail.com")
tags_following = kakar.relation.follows_tag.all()
This is fine.
But, to access intermediate fields I have to go through a big list of other queries. Suppose I want to display when the user started following a tag, I will have to do something like this:
kakar = CustomUser.objects.get(email="kakar#gmail.com")
kakar_relation = Relation.objects.get(user=kakar)
t1 = kakar.relation.follows_tag.all()[0]
kakar_t1_relation = TagRelation.objects.get(user=kakar_relation, following_tag=t1)
kakar_t1_relation.pub_date
As you can see, just to get the date I have to go through so much query. Is this the only way to get intermediate values, or this can be optimized? Also, I am not sure if this model design is the way to go, so if you have any recomendation or advice I would be very grateful. Thank you.
You need to use Double underscore i.e. ( __ ) for ForeignKey lookup,
Like this :
user_tags = TagRelation.objects.filter(user__user__email="kakar#gmail.com").values("following_tag__name", "pub_date")
If you need the name of the tag, you can use following_tag__name in the query and if you need id you can use following_tag__id.
And for that you need to iterate through the result of above query set, like this:
for items in user_tags:
print items['following_tag__name']
print items['pub_date']
One more thing,The key word values will return a list of dictionaries and you can iterate it through above method and if you are using values_list in the place of values, it will return a list of tuples. Read further from here .

Django - Update multiple integer fields at once

I'd like to update multiple integer fields at once in following model.
class Foo(models.Model):
field_a = models.PositiveIntegerField()
field_b = models.PositiveIntegerField()
field_c = models.PositiveIntegerField()
Originally, it can be done like following code with two queries.
foo = Foo.objects.get(id=1)
foo.field_a += 1
foo.field_b -= 1
foo.field_c += 2
foo.save()
I'd like make it more simpler with update in one query.
However, following attempts raised error.
# 1st attempt
Foo.objects.filter(id=1).update(
field_a=F('field_a')+1,
field_b=F('field_a')-1,
field_c=F('field_a')+2)
# 2nd attempt
Foo.objects.filter(id=1).\
update(field_a=F('field_a')+1).\
update(field_b=F('field_b')-1) ).\
update(field_c=F('field_c')+2)
How can I solve this ?
Form the django docs:
Calls to update can also use F expressions to update one field based on the value of another field in the model. This is especially useful for incrementing counters based upon their current value. For example, to increment the pingback count for every entry in the blog:
>>> from django.db.models import F
>>> Entry.objects.all().update(n_pingbacks=F('n_pingbacks') + 1)
You have to have an instance of Foo or a queryset before you can update. You should do something like this:
Foo.objects.get(id=1)update(field_a=F('field_a')+1).\
update(field_b=F('field_b')-1) ).\
update(field_c=F('field_c')+2)
or
Foo.objects.filter(id__in=[1,3,6,7]).update(field_a=F('field_a')+1).\
update(field_b=F('field_b')-1) ).\
update(field_c=F('field_c')+2)
Reference: https://docs.djangoproject.com/en/1.8/topics/db/queries/#updating-multiple-objects-at-once
If save() is passed a list of field names in keyword argument update_fields, only the fields named in that list will be updated. This may be desirable if you want to update just one or a few fields on an object. There will be a slight performance benefit from preventing all of the model fields from being updated in the database. For example:
product.name = 'Name changed again'
product.save(update_fields=['name'])
see more docs [here]:https://docs.djangoproject.com/en/dev/ref/models/instances/#specifying-which-fields-to-save

Django - sorting object based on user defined order in template

I want the user to be able to order a list of objects in a table using javascript. Then, in a django function I would like to sort those object based on the same ordering, not on an attribute.
Is it possible? I was thinking about passing a list of pk from the template to the view and then ordering the objects according to this list, but I have not found a way to do it yet.
I don't think this is possible with queryset. Try following:
pk_list = [2, 1, 3, 4]
pk2obj = {obj.pk: obj for obj in Model.objects.all()}
objects_ordered = [pk2obj[pk] for pk in pk_list]
pkg2obj is mapping between pk and model instance object. To make a dictionary I used dictionary comprehension.
If you want to omit deleted objects:
objects_ordered = [pk2obj[pk] for pk in pk_list if pk in pk2obj]
Else if you want to replace deleted objects with default value (None in following code):
objects_ordered = [pk2obj.get(pk, None) for pk in pk_list]
I've had to solve this exact problem before.
If you want the user to be able to reorder them into a user-defined order, you can easily define a field to store this order.
As you say, initially, you could serve them in order according to id or an upload_date DateTimeField. But you could also have an PositiveIntegerField in the model, named position or order, to represent the user-defined order.
class MediaItem(models.Model):
user = models.ForeignKey(User)
upload_date = models.DateTimeField(auto_now_add = True)
position = models.PositiveIntegerField()
Whenever a user changes the order on the frontend, the JS can send the new order as an array of objects (ie. new_order = [{"pk":3, "position":1}, {"pk":1, "position":2}, {"pk":2, "position":3}]). The view can look up each instance by pk, and change the position:
for obj in new_order:
media_item = MediaItem.objects.get(pk=obj['pk'])
media_item.position = obj['position']
media_item.save()
Then always query using
objects_ordered.objects.order_by('position')
That's how we managed to do it. If you have more specific questions regarding this approach, feel free to ask in the comments.
Edit:
If the same object can be a member of many different groups or lists, and you want to store the position of the membership within that list, you can achieve this using a through model. A through model is useful when you need to store data that relates to the relationship between two objects that are related. In addition to the MediaItem class shown above, this is what your other models would look like:
class Album(models.Model):
media_items = models.ManyToManyField(MediaItem,
related_name = 'album_media_items',
through = 'Membership')
class Membership(models.Model):
album = models.ForeignKey(Album,
related_name = 'album')
media_item = models.ForeignKey(MediaItem,
related_name = 'media_item')
date = models.DateTimeField(auto_now_add = True)
position = models.PositiveIntegerField()
Then, you could query the Membership instances, instead of the MediaItem instances.
# get id of list, or album...
alb = Album.objects.get(pk=id_of_album)
media_items = Membership.objects.filter(album=alb).order_by('position')
for item in media_items:
# return the items, or do whatever...
# keep in mind they are now in the user-defined order
You can do this:
pk_list = [1,5,3,9]
foo = Foo.objects.filter(id__in=pk_list)
#Order your QuerySet in base of your pk_list using Lambda
order_foo = sorted(foo, key = lambda:x , pk_list.index(x.pk))