Django Class 基于视图：覆盖表单名称

Question

我是 Django 的新手。我尝试构建一个基于 class 的视图以创建一个对象。

模板中表单的默认名称是form，我想将其更改为"ajoutersource"，但我不知道如何更改。

views.py

class ajoutSource(CreateView):
    model = Source
    template_name = "pmd/ajouterSource.html"
    form_class = AjouterSourceForm
    success_url = reverse_lazy(ListerSources)

ajouterSource.html

{% for field in ajoutersource %} 
    <div class="row"> 
        {% if field.errors %}
            <div class="error">{{ field.errors }}</div> 
        {% endif %}
        <div class="label">{{ field.label }}</div> 
        <div class="field">{{ field }}</div>
    </div> 
{% endfor %}

Answer 1

覆盖 get_context_data():

class ajoutSource(CreateView):
    model = Source
    template_name = "pmd/ajouterSource.html"
    form_class = AjouterSourceForm
    success_url = reverse_lazy(ListerSources)

    def get_context_data(self, **kwargs):
        context = super(ajoutSource, self).get_context_data(**kwargs)
        context["ajoutersource"]=context["form"]
        return context

Answer 2

按照下面的方法就可以了

方法一（模型形式）

def get_context_data(self, **kwargs):
    context = super().get_context_data(**kwargs)
    context['new_name'] = self.get_form()
    return context

方法 2（简单形式）

def get_context_data(self, **kwargs):
    context = super().get_context_data(**kwargs)
    context['new_name'] = context["form"]
    return context

推荐方法一 （注意：这是 python 3.6+ 语法，更改 super() 调用 python 2.0+）

Answer 3

覆盖 get_context_data，context['form'] 取自 SomeForm，并更改为 form_1，您可以在模板中使用 form_1

class Something(generic.CreateView):
    template_name = 'app/example.html'
    form_class = forms.SomeForm
    model = models.SomeModel
        
    def get_context_data(self, **kwargs):
        context = super(Something, self).get_context_data(**kwargs)
        context["form_1"] = context["form"]
        context["form_2"] = forms.SomeForm2(**self.get_form_kwargs())
        return context

Django Class 基于视图：覆盖表单名称

Django Class Based Views : Override form name

python

django

django-generic-views

django-class-based-views