python - 如何在 Django View 之间传递数据

Question

This问题一般解决我的问题，但我正在寻找更具体的解释。

我希望用户更新一组模型对象，但是，需要先检索这些对象的查询集。我的计划是在两个单独的 UR/ View 中执行此操作，从第一个获取查询集信息，然后显示接下来要更新的模型表单集。

我的第一个 View 给出了所有“项目”的列表(我的模型之一)，并检索了所选项目的 ID。

这是表格:

class ProjectLookupForm(forms.Form):
    Project_Name = chosenforms.ChosenModelChoiceField(queryset=Project.objects.all())

这是 View :

def update_project_filter(request):
    project_form = ProjectLookupForm(request.POST or None)
    if request.method == 'POST':
        if project_form.is_valid():
            context = {"project_form":project_form}
            # Get project here and share it with the next view.
            selected_project_id = project_form.cleaned_data["Project_Name"].id
            # Add a new return statement here?
            # Or call update project view from here?
            # Add a redirect button to html?
        else:
            errors = project_form.errors
            context = {"errors":errors, "project_form":project_form}
    else:
        context = {"project_form":project_form}
    return render(request, 'filter_update_project_form.html', context)

如您所见，我已经包含了一些集思广益我的可能性的评论。我的目标是将 selected_project_id 发送到下一个 View ，以便它可以将该 id 用作模型表单查询集。

def update_project(request):
    UpdateFormset = modelformset_factory(Sample, fields=("sample_name", "extraction_date", 
                                                     "project", "order", "notebook", "notebook_page"))
    if request.method == 'POST':
        formset = UpdateFormset(request.POST, request.FILES)
        if formset.is_valid():
            formset.save()
            context = {"formset": formset, "project_form":project_form}
        else:
            errors = formset.errors
            context = {"formset":formset, "errors":errors, "project_form":project_form}
    else:
        formset = UpdateFormset(queryset=Sample.objects.filter(project=2))
        context = {"formset":formset, "project_form":project_form}
    return render(request, 'update_project_form.html', context)

在这里可以看到我已经像这样对查询集进行了硬编码:

queryset=Sample.objects.filter(project=2)

如何将“project=”设置为我的 selected_project_id？我是否将此信息作为输入参数传递给 View ？还是我将它发送到下一个 URL 并从那里获取？

Answer 1

假设您已激活 django.contrib.sessions.middleware.SessionMiddleware ;您可以使用 request.session 字典在 View 之间传递数据，如下所示:

def update_project_filter(request):
    ...
    selected_project_id = project_form.cleaned_data["Project_Name"].id
    request.session['selected_project_id'] = selected_project_id
    ...

def update_project(request):
    ...
    selected_project_id = request.session.get('selected_project_id')
    ...