I have a model formset that I want to display 10 forms at a time using Django\'s Paginator, but it can\'t be done like paginator = Paginator(formset, 10)
. What\
A more elegant solution is to set ordered=True
on the Page object so that it can be passed to a ModelFormSet
.
Here is an example:
forms_per_page = 10
current_page = 1
ModelFormSet = modelformset_factory(MyModel, form=MyForm)
queryset = MyModel.objects.all()
paginator = Paginator(queryset, forms_per_page)
page_object = paginator.page(current_page)
page_object.ordered = True
form = ModelFormSet(queryset=page_object)
This is more efficient than the accepted answer because avoids the second database query that takes place in the line:
page_query = query.filter(id__in=[object.id for object in objects])
The problem here is that you're using brands (a Page
) in a context that's expecting a QuerySet
. So, we need that damn QuerySet
. You are in right way, but a lot of code.
In source code we have:
class Page(collections.Sequence):
def __init__(self, object_list, number, paginator):
self.object_list = object_list
self.number = number
self.paginator = paginator
...
So, our queryset in self.object_list
attribute and just use it!
formset = SomeModelFormSet(queryset=objects.object_list)
More correct way to use this
...
formset = FormSet(queryset=page_query.object_list)
...
This is a generic example of the solution I found to my problem:
In the forms.py
file:
class MyForm(ModelForm):
class Meta:
model = MyModel
fields = ('description',)
In the views.py
file:
from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger
FormSet = modelformset_factory(MyModel, form=MyForm, extra=0)
if request.method == 'POST':
formset = FormSet(request.POST, request.FILES)
# Your validation and rest of the 'POST' code
else:
query = MyModel.objects.filter(condition)
paginator = Paginator(query, 10) # Show 10 forms per page
page = request.GET.get('page')
try:
objects = paginator.page(page)
except PageNotAnInteger:
objects = paginator.page(1)
except EmptyPage:
objects = paginator.page(paginator.num_pages)
page_query = query.filter(id__in=[object.id for object in objects])
formset = FormSet(queryset=page_query)
context = {'objects': objects, 'formset': formset}
return render_to_response('template.html', context,
context_instance=RequestContext(request))
You need to create the formset with the objects in the present page, otherwise, when you try to do formset = FormSet(request.POST, request.FILES)
in the POST method, Django raises a MultiValueDictKeyError error.
In the template.html
file:
{% if objects %}
<form action="" method="post">
{% csrf_token %}
{{ formset.management_form }}
{% for form in formset.forms %}
{{ form.id }}
<!-- Display each form -->
{{ form.as_p }}
{% endfor %}
<input type="submit" value="Save" />
</form>
<div class="pagination">
<span class="step-links">
{% if objects.has_previous %}
<a href="?page={{ objects.previous_page_number }}">Previous</a>
{% endif %}
<span class="current">
Page {{ objects.number }} of {{ objects.paginator.num_pages }}
</span>
{% if objects.has_next %}
<a href="?page={{ objects.next_page_number }}">next</a>
{% endif %}
</span>
</div>
{% else %}
<p>There are no objects.</p>
{% endif %}
Agree with Elrond Supports Monica. Fake attribute is interesting way to resolve the ordering error (Cannot reorder a query once a slice has been taken.)
But it can be fixed in one line also
queryset = queryset.order_by(Entry._meta.pk.name)
This fake ordering is need for avoid error in django.form.modelsBaseModelFormSet(BaseFormSet).get_queryset(): line #640
that make artificial ordering by pk but it impossible after slicing
(LIMIT-ations in SQL )
More detailed example
queryset = Entry.objects.all()
queryset = queryset.order_by(Entry._meta.pk.name)
paginator = Paginator(object_list=queryset, per_page=10)
page_obj = paginator.get_page(request.GET.get('page'))
EntryFormSet = modelformset_factory(Entry, EntryForm, extra=0)
entryformset = EntryFormSet(queryset=page_obj.object_list)