Paginate Django formset

Question

我有一個模型formset，我想一次使用Django的Paginator顯示10個表單，但它不能像paginator = Paginator(formset, 10)那樣完成。如果有辦法，那麼做到這一點的正確方法是什麼？

Answer 1

這是我發現我的問題的解決方案的通用例子：

在forms.py文件：

class MyForm(ModelForm): 
    class Meta: 
     model = MyModel 
     fields = ('description',) 

在views.py文件：

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)) 

您需要創建在當前頁面中包含對象的formset，否則，當您在POST方法中嘗試執行formset = FormSet(request.POST, request.FILES)時，Django引發MultiValueDictKeyError錯誤。

在template.html文件：

{% 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 %} 

Answer 2

更正確的方式來使用這個

... 
formset = FormSet(queryset=page_query.object_list) 
... 

Answer 3

在這裏的問題是，您使用的上下文品牌（一個Page）這是期待QuerySet 。所以，我們需要那該死的QuerySet。你是正確的，但很多代碼。

在源代碼中，我們有：

class Page(collections.Sequence): 

    def __init__(self, object_list, number, paginator): 
     self.object_list = object_list 
     self.number = number 
     self.paginator = paginator 
     ... 

所以，我們在self.object_list屬性查詢集，只是用它！

formset = SomeModelFormSet(queryset=objects.object_list)