Django – Getting last object created, simultaneous filters
Question:
Apologies, I am completely new to Django and Python.
I have 2 questions. First, how would I go about getting the last object created (or highest pk) in a list of objects? For example, I know that I could use the following to get the first object:
list = List.objects.all()[0]
Is there a way to get the length of List.objects? I’ve tried List.objects.length but to no avail.
Second, is it possible to create simultaneous filters or combine lists? Here is an example:
def findNumber(request, number)
phone_list = Numbers.objects.filter(cell=number)
I want something like the above, but more like:
def findNumber(request, number)
phone_list = Numbers.objects.filter(cell=number or home_phone=number)
What is the correct syntax, if any?
Answers:
You can use the count() method on a query set the get the number of items.
list = List.objects.all()
list.count()
Arguments to filter are “AND”ed together. If you need to do OR filters look at Q objects.
http://docs.djangoproject.com/en/dev/topics/db/queries/#complex-lookups-with-q-objects
I haven’t tried this yet, but I’d look at the latest() operator on QuerySets:
latest(field_name=None)
Returns the latest object in the
table, by date, using the field_name
provided as the date field.
This example returns the latest Entry
in the table, according to the
pub_date field:
Entry.objects.latest(‘pub_date’)
If your model’s Meta specifies
get_latest_by, you can leave off the
field_name argument to latest().
Django will use the field specified in
get_latest_by by default.
Like get(), latest() raises
DoesNotExist if an object doesn’t
exist with the given parameters.
Note latest() exists purely for
convenience and readability.
And the model docs on get_latest_by:
get_latest_by
Options.get_latest_by
The name of a DateField or DateTimeField in the model. This specifies the default field to use in your model Manager’s latest method.
Example:
get_latest_by = “order_date”
See the docs for latest() for more.
Edit: Wade has a good answer on Q() operator.
For the largest primary key, try this:
List.objects.order_by('-pk')[0]
Note that using pk works regardless of the actual name of the field defined as your primary key.
this works!
Model.objects.latest('field') – field can be id. that will be the latest id
alternative for the latest object created:
List.objects.all()[List.objects.count()-1]
It is necessary to add an AssertionError for the case when there are no items in the list.
except AssertionError:
...
Since Django 1.6 – last
last()
Works like first(), but returns the last object in the queryset.
Returns the last object matched by the queryset, or None if there is no matching object. If the QuerySet has no ordering defined, then the queryset is automatically ordered by the primary key.
list = List.objects.last() gives you the last object created
I am working on Django version is 1.4.22, neither last nor lastet is working.
My way to solve with minimum db loading is like:
latest = lambda model_objects, field : model_objects.values_list( field, flat = True ).order_by( "-" + field )[ 0 ]
latest_pk = latest( List.objects, "pk" )
This function accepts query_set as input.
You may bind this function dynamically by doing:
import types
List.objects.latest = types.MethodType( latest, List.objects )
Then you should be able to get the last object by this latest pk easily.
Try this:
InsertId= (TableName.objects.last()).id
Apologies, I am completely new to Django and Python.
I have 2 questions. First, how would I go about getting the last object created (or highest pk) in a list of objects? For example, I know that I could use the following to get the first object:
list = List.objects.all()[0]
Is there a way to get the length of List.objects? I’ve tried List.objects.length but to no avail.
Second, is it possible to create simultaneous filters or combine lists? Here is an example:
def findNumber(request, number)
phone_list = Numbers.objects.filter(cell=number)
I want something like the above, but more like:
def findNumber(request, number)
phone_list = Numbers.objects.filter(cell=number or home_phone=number)
What is the correct syntax, if any?
You can use the count() method on a query set the get the number of items.
list = List.objects.all()
list.count()
Arguments to filter are “AND”ed together. If you need to do OR filters look at Q objects.
http://docs.djangoproject.com/en/dev/topics/db/queries/#complex-lookups-with-q-objects
I haven’t tried this yet, but I’d look at the latest() operator on QuerySets:
latest(field_name=None)
Returns the latest object in the
table, by date, using the field_name
provided as the date field.This example returns the latest Entry
in the table, according to the
pub_date field:Entry.objects.latest(‘pub_date’)
If your model’s Meta specifies
get_latest_by, you can leave off the
field_name argument to latest().
Django will use the field specified in
get_latest_by by default.Like get(), latest() raises
DoesNotExist if an object doesn’t
exist with the given parameters.Note latest() exists purely for
convenience and readability.
And the model docs on get_latest_by:
get_latest_by
Options.get_latest_by
The name of a DateField or DateTimeField in the model. This specifies the default field to use in your model Manager’s latest method.
Example:
get_latest_by = “order_date”
See the docs for latest() for more.
Edit: Wade has a good answer on Q() operator.
For the largest primary key, try this:
List.objects.order_by('-pk')[0]
Note that using pk works regardless of the actual name of the field defined as your primary key.
this works!
Model.objects.latest('field') – field can be id. that will be the latest id
alternative for the latest object created:
List.objects.all()[List.objects.count()-1]
It is necessary to add an AssertionError for the case when there are no items in the list.
except AssertionError:
...
Since Django 1.6 – last
last()
Works like first(), but returns the last object in the queryset.
Returns the last object matched by the queryset, or None if there is no matching object. If the QuerySet has no ordering defined, then the queryset is automatically ordered by the primary key.
list = List.objects.last() gives you the last object created
I am working on Django version is 1.4.22, neither last nor lastet is working.
My way to solve with minimum db loading is like:
latest = lambda model_objects, field : model_objects.values_list( field, flat = True ).order_by( "-" + field )[ 0 ]
latest_pk = latest( List.objects, "pk" )
This function accepts query_set as input.
You may bind this function dynamically by doing:
import types
List.objects.latest = types.MethodType( latest, List.objects )
Then you should be able to get the last object by this latest pk easily.
Try this:
InsertId= (TableName.objects.last()).id