How to change the status of the model using the button?

Question

Django

There is a model with two statuses

class Post(models.Model):
STATUS_CHOICES = (
        ('draft', 'Draft'),
        ('published', 'Published'),
    )
    title = models.CharField(_('Покупатель'), max_length=50)
    slug = models.SlugField(unique=True, null=True, blank=True)
    status = models.CharField(max_length=10,
                              choices=STATUS_CHOICES,
                              default='published')

On the single post output page, how to add a button to change the post status?

class DetailPageView(View):
    def get(self, request, pk):
        post = Post.objects.get(id=pk)
        return render(request, 'updocks/adddoc/detail.html', {'post': post})

{% block content %}
    <h1>{{ post.title }}</h1>
    {{ post.body|linebreaks }}
{% endblock %}