I want to do pretty much the same like in this ticket at djangoproject.com, but with some additonal formatting. From this query
>>> MyModel.objects.values('cryptic_value_name')
[{'cryptic_value_name': 1}, {'cryptic_value_name': 2}]
I want to get something like that:
>>> MyModel.objects.values(renamed_value='cryptic_value_name')
[{'renamed_value': 1}, {'renamed_value': 2}]
Is there another, more builtin way or do I have to do this manually?
From django>=1.8 you can use annotate and F object
from django.db.models import F
MyModel.objects.annotate(renamed_value=F('cryptic_value_name')).values('renamed_value')
Also extra() is going to be deprecated, from the django docs:
This is an old API that we aim to deprecate at some point in the future. Use it only if you cannot express your query using other queryset methods. If you do need to use it, please file a ticket using the QuerySet.extra keyword with your use case (please check the list of existing tickets first) so that we can enhance the QuerySet API to allow removing extra(). We are no longer improving or fixing bugs for this method.
Without using any other manager method (tested on v3.0.4):
from django.db.models import F
MyModel.objects.values(renamed_value=F('cryptic_value_name'))
Excerpt from Django docs:
An F() object represents the value of a model field or annotated
column. It makes it possible to refer to model field values and
perform database operations using them without actually having to pull
them out of the database into Python memory.
It's a bit hacky, but you could use the extra method:
MyModel.objects.extra(
select={
'renamed_value': 'cryptic_value_name'
}
).values(
'renamed_value'
)
This basically does SELECT cryptic_value_name AS renamed_value in the SQL.
Another option, if you always want the renamed version but the db has the cryptic name, is to name your field with the new name but use db_column to refer to the original name in the db.
I am working with django 1.11.6
( And the key:value pair is opposite to that of accepted answer )
This is how i am making it work for my project
def json(university):
address = UniversityAddress.objects.filter(university=university)
address = address.extra(select={'city__state__country__name': 'country', 'city__state__name': 'state', 'city__name': 'city'})
address = address.values('country', 'state', "city", 'street', "postal_code").get()
return address
Note that adding simultanous objects.filter().extra().values() is same as above.
Try passing as kwargs:
MyModel.objects.annotate(**{'A B C':F('profile_id')}).values('A B C')
In my case, there were spaces and other special characters included in the key of each value in the result set so this did the trick.
Its more than simple if you want to rename few fields of the mode.
Try
projects = Project.objects.filter()
projects = [{'id': p.id, 'name': '%s (ID:%s)' % (p.department, p.id)} for p in projects]
Here i do not have a name field in the table, but i can get that after tweaking a little bit.
Related
I have a filter which should return a queryset with 2 objects, and should have one different field. for example:
obj_1 = (name='John', age='23', is_fielder=True)
obj_2 = (name='John', age='23', is_fielder=False)
Both the objects are of same model, but different primary key. I tried usign the below filter:
qs = Model.objects.filter(name='John', age='23').annotate(is_fielder=F('plays__outdoor_game_role')=='Fielder')
I used annotate first time, but it gave me the below error:
TypeError: QuerySet.annotate() received non-expression(s): False.
I am new to Django, so what am I doing wrong, and what should be the annotate to get the required objects as shown above?
The solution by #ktowen works well, quite straightforward.
Here is another solution I am using, hope it is helpful too.
queryset = queryset.annotate(is_fielder=ExpressionWrapper(
Q(plays__outdoor_game_role='Fielder'),
output_field=BooleanField(),
),)
Here are some explanations for those who are not familiar with Django ORM:
Annotate make a new column/field on the fly, in this case, is_fielder. This means you do not have a field named is_fielder in your model while you can use it like plays.outdor_game_role.is_fielder after you add this 'annotation'. Annotate is extremely useful and flexible, can be combined with almost every other expression, should be a MUST-KNOWN method in Django ORM.
ExpressionWrapper basically gives you space to wrap a more complecated combination of conditions, use in a format like ExpressionWrapper(expression, output_field). It is useful when you are combining different types of fields or want to specify an output type since Django cannot tell automatically.
Q object is a frequently used expression to specify a condition, I think the most powerful part is that it is possible to chain the conditions:
AND (&): filter(Q(condition1) & Q(condition2))
OR (|): filter(Q(condition1) | Q(condition2))
Negative(~): filter(~Q(condition))
It is possible to use Q with normal conditions like below:
(Q(condition1)|id__in=[list])
The point is Q object must come to the first or it will not work.
Case When(then) can be simply explained as if con1 elif con2 elif con3 .... It is quite powerful and personally, I love to use this to customize an ordering object for a queryset.
For example, you need to return a queryset of watch history items, and those must be in an order of watching by the user. You can do it with for loop to keep the order but this will generate plenty of similar queries. A more elegant way with Case When would be:
item_ids = [list]
ordering = Case(*[When(pk=pk, then=pos)
for pos, pk in enumerate(item_ids)])
watch_history = Item.objects.filter(id__in=item_ids)\
.order_by(ordering)
As you can see, by using Case When(then) it is possible to bind those very concrete relations, which could be considered as 1) a pinpoint/precise condition expression and 2) especially useful in a sequential multiple conditions case.
You can use Case/When with annotate
from django.db.models import Case, BooleanField, Value, When
Model.objects.filter(name='John', age='23').annotate(
is_fielder=Case(
When(plays__outdoor_game_role='Fielder', then=Value(True)),
default=Value(False),
output_field=BooleanField(),
),
)
I have these two models:
class A(models.Model):
name=models.CharField(max_length=10)
class D(models.Model):
code=models.IntegerField()
the code field can have a number that exists in model A but it cant be related due to other factors. But what I want know is to list items from A whose value is the same with code
items=D.objects.values('code__name')
would work but since they are not related nor can be related, how can I handle that?
You can use Subquery() expressions in Django 1.11 or newer.
from django.db.models import OuterRef, Subquery
code_subquery = A.objects.filter(id=OuterRef('code'))
qs = D.objects.annotate(code_name=Subquery(code_subquery.values('name')))
The output of qs is a queryset of objects D with an added field code_name.
Footnotes:
It is compiled to a very similar SQL (like the Bear Brown's solution with "extra" method, but without disadvantages of his solution, see there):
SELECT app_d.id, app_d.code,
(SELECT U0.name FROM app_a U0 WHERE U0.id = (app_d.code)) AS code_name
FROM app_d
If a dictionary output is required it can be converted by .values() finally. It can work like a left join i.e. if the pseudo related field allows null (code = models.IntegerField(none=True)) then the objects D are not restricted and the output code_name value could be None. A feature of Subquery is that it returns only one field expression must be eventually repeated for another fields. (That is similar to extra(select={...: "SELECT ..."}), but thanks to object syntax it can be more readable customized than an explicit SQL.)
you can use django extra, replace YOUAPP on your real app name
D.objects.extra(select={'a_name': 'select name from YOUAPP_a where id=code'}).values('a_name')
# Replace YOUAPP^^^^^
I'd like to update a table with Django - something like this in raw SQL:
update tbl_name set name = 'foo' where name = 'bar'
My first result is something like this - but that's nasty, isn't it?
list = ModelClass.objects.filter(name = 'bar')
for obj in list:
obj.name = 'foo'
obj.save()
Is there a more elegant way?
Update:
Django 2.2 version now has a bulk_update.
Old answer:
Refer to the following django documentation section
Updating multiple objects at once
In short you should be able to use:
ModelClass.objects.filter(name='bar').update(name="foo")
You can also use F objects to do things like incrementing rows:
from django.db.models import F
Entry.objects.all().update(n_pingbacks=F('n_pingbacks') + 1)
See the documentation.
However, note that:
This won't use ModelClass.save method (so if you have some logic inside it won't be triggered).
No django signals will be emitted.
You can't perform an .update() on a sliced QuerySet, it must be on an original QuerySet so you'll need to lean on the .filter() and .exclude() methods.
Consider using django-bulk-update found here on GitHub.
Install: pip install django-bulk-update
Implement: (code taken directly from projects ReadMe file)
from bulk_update.helper import bulk_update
random_names = ['Walter', 'The Dude', 'Donny', 'Jesus']
people = Person.objects.all()
for person in people:
r = random.randrange(4)
person.name = random_names[r]
bulk_update(people) # updates all columns using the default db
Update: As Marc points out in the comments this is not suitable for updating thousands of rows at once. Though it is suitable for smaller batches 10's to 100's. The size of the batch that is right for you depends on your CPU and query complexity. This tool is more like a wheel barrow than a dump truck.
Django 2.2 version now has a bulk_update method (release notes).
https://docs.djangoproject.com/en/stable/ref/models/querysets/#bulk-update
Example:
# get a pk: record dictionary of existing records
updates = YourModel.objects.filter(...).in_bulk()
....
# do something with the updates dict
....
if hasattr(YourModel.objects, 'bulk_update') and updates:
# Use the new method
YourModel.objects.bulk_update(updates.values(), [list the fields to update], batch_size=100)
else:
# The old & slow way
with transaction.atomic():
for obj in updates.values():
obj.save(update_fields=[list the fields to update])
If you want to set the same value on a collection of rows, you can use the update() method combined with any query term to update all rows in one query:
some_list = ModelClass.objects.filter(some condition).values('id')
ModelClass.objects.filter(pk__in=some_list).update(foo=bar)
If you want to update a collection of rows with different values depending on some condition, you can in best case batch the updates according to values. Let's say you have 1000 rows where you want to set a column to one of X values, then you could prepare the batches beforehand and then only run X update-queries (each essentially having the form of the first example above) + the initial SELECT-query.
If every row requires a unique value there is no way to avoid one query per update. Perhaps look into other architectures like CQRS/Event sourcing if you need performance in this latter case.
Here is a useful content which i found in internet regarding the above question
https://www.sankalpjonna.com/learn-django/running-a-bulk-update-with-django
The inefficient way
model_qs= ModelClass.objects.filter(name = 'bar')
for obj in model_qs:
obj.name = 'foo'
obj.save()
The efficient way
ModelClass.objects.filter(name = 'bar').update(name="foo") # for single value 'foo' or add loop
Using bulk_update
update_list = []
model_qs= ModelClass.objects.filter(name = 'bar')
for model_obj in model_qs:
model_obj.name = "foo" # Or what ever the value is for simplicty im providing foo only
update_list.append(model_obj)
ModelClass.objects.bulk_update(update_list,['name'])
Using an atomic transaction
from django.db import transaction
with transaction.atomic():
model_qs = ModelClass.objects.filter(name = 'bar')
for obj in model_qs:
ModelClass.objects.filter(name = 'bar').update(name="foo")
Any Up Votes ? Thanks in advance : Thank you for keep an attention ;)
To update with same value we can simply use this
ModelClass.objects.filter(name = 'bar').update(name='foo')
To update with different values
ob_list = ModelClass.objects.filter(name = 'bar')
obj_to_be_update = []
for obj in obj_list:
obj.name = "Dear "+obj.name
obj_to_be_update.append(obj)
ModelClass.objects.bulk_update(obj_to_be_update, ['name'], batch_size=1000)
It won't trigger save signal every time instead we keep all the objects to be updated on the list and trigger update signal at once.
IT returns number of objects are updated in table.
update_counts = ModelClass.objects.filter(name='bar').update(name="foo")
You can refer this link to get more information on bulk update and create.
Bulk update and Create
class Log:
project = ForeignKey(Project)
msg = CharField(...)
date = DateField(...)
I want to select the four most recent Log entries where each Log entry must have a unique project foreign key. I've tried the solutions on google search but none of them works and the django documentation isn't that very good for lookup..
I tried:
id_list = Log.objects.order_by('-date').values_list('project_id').distinct()[:4]
entries = Log.objects.filter(id__in=id_list)
id_list is empty unless I remove the order_by() but then it's not in the correct order.
entries = Log.objects.filter(id__in=id_list)
The objects in entries is not in the same order as in id_list because when you use Mysql function IN() it will not sort the result by the input order ... How to do it in django?
It looks like it is impossible to achieve what you want with django orm. Documentation states that is not possible to use order_by along with distinct.
However there might be another way to solve it. Maybe you could select Project objects, and annotate them with latest log entries.
Here's a single-query solution (but it will probably be too slow):
Log.objects.filter(project__log__date__gte=F('date')).annotate(c=Count('project__log')).filter(c__lte=4).order_by('project', 'c')
I think that Skirmantas is right and you have to do it in a more complex way:
from django.db.models import Max
projects = Project.objects.annotate(last_logged=Max('log__date')).order_by('-last_logged')[:4]
log_entries = [proj.log_set.order_by('-date')[0] for proj in projects]
Should entry_set be cached with select_related? My DB is still getting calls even after I use select_related. The pertinent sections
class Alias(models.Model):
achievements = models.ManyToManyField('Achievement', through='Achiever')
def points(self) :
points = 0
for a in self.achiever_set.all() :
points += a.achievement.points * a.count
return points
class Achievement(models.Model):
name = models.CharField(max_length=100)
points = models.IntegerField(default=1)
class Achiever(models.Model):
achievement = models.ForeignKey(Achievement)
alias = models.ForeignKey(Alias)
count = models.IntegerField(default=1)
aliases = Alias.objects.all().select_related()
for alias in aliases :
print "points : %s" % alias.points()
for a in alias.achiever_set.all()[:5] :
print "%s x %d" % (a.achievement.name, a.count)
And I'm seeing a big join query at the start, and then individual calls for each achievement. Both for the points and for the name lookup.
Is this a bug, or am I doing something wrong?
With Django 1.4 you can use prefetch_related which will work for ManyToMany relations:
https://docs.djangoproject.com/en/dev/ref/models/querysets/#prefetch-related
Select_related() doesn't work with manytomanyfields. At the moment, this is something that is not planned, but might be a future feature. See http://code.djangoproject.com/ticket/6432
In this case, if you want to make a single query you got two options
1) Make your own SQL, probably won't be pretty or fast.
2) You could also query on the model with the foreignkey. You would be able to use select_related in that case. You stil won't be able to access the modelname_set but with some formatting you would be able to vet the data you need in a single query. None of the options are ideal, but you could get it working at a deacent speed aswell.
In Django 1.3 You can use Queryset.values() and do something like:
Alias.objects[.filter().exclude() etc.].values('achievements__name', 'achievement__points')
Only drwaback is that You get QuerySetList instead of QuerySet. But this can be simply overcome by passing all necessary fields into values() - You have to change Your perception ;)
This can save you few dosen of queries...
Details can be found here in django docs:
http://docs.djangoproject.com/en/dev/ref/models/querysets/#django.db.models.query.QuerySet.values