Django编辑Froala窗体的视图/模板

Question

我想使用弗洛拉编辑 (或类似的编辑器)在我的网站上创建和编辑文章。我可以用这个表单创建任何文章/可以从编辑器获得数据，但是我不知道如何通过视图函数将来自DB的文章数据插入到模板中编辑器的{{ form }}中。怎么做？在这种情况下，是否还有其他更合适的表单/编辑器？

forms.py

from froala_editor.widgets import FroalaEditor
    
class FroalaModelForm(forms.ModelForm):
        content = forms.CharField(widget=FroalaEditor)
    
        class Meta:
            model = FroalaModel
            fields = ('name', 'content')

views.py

def post_froala_form(request):
    form = FroalaModelForm(request.POST)
    if form.is_valid():
        form = form.save(commit=False)
        form.name = request.POST['name']
        form.content = request.POST['content']
        form.save()
        return HttpResponseRedirect(reverse('polls:froala_index'))
    else:
        return render(request, 'polls/froala template.html', {
            'form': form
        })

froala_template.html

{% load crispy_forms_filters %}
{% load crispy_forms_tags %}

{% block title %}
    Froala template
{% endblock %}

{% block content %}
    <form action="{% url 'polls:post_froala_form' %}" method="post">
        {% csrf_token %}
        {{ form.media }}
        {{ form|crispy }}
        <button type="submit">Send form</button>
    </form>
{% endblock content %}

Answer 1

我用Django表单决定了这一点，在参数"instance“中从数据库中指示了一个字段：

views.py

def question_create_update_form(request, question_id=None):
    if question_id:
        question = Question.objects.get(id=question_id)
        if question.author_name == request.user.username:
            form = AddQuestionForm(instance=question)
            return render(request, 'polls/add_question.html', {
                'title': question.question_title,
                'choices': question.choice_set.all(),
                'tags': question.tag.all(),
                'question_id': question_id,
                'form': form

            })
        else:
            return HttpResponseRedirect(reverse('polls:create_question'))
    else:
        return render(request, 'polls/add_question.html', {
            'form': AddQuestionForm
        })