How to exclude django form field from render but still keep it for validation?

Question

If I have a model (extract):

class Coupon(models.Model):
    image = models.ImageField(max_length=64, null=True, upload_to=\"couponimages\")

<

Answer 1

You can render form manualy like this:

in views.py

return render(request, 'template.html', {'form': form})

in template.html

<form action="/your-name/" method="post" enctype="multipart/form-data">
    {% csrf_token %}
    {{ form.field1 }}
    {{ form.field2 }}
          .
          .
          .
    <input type="submit" value="Submit" />
</form>

also take a look to Working with forms.

Answer 2

I think the best way to achieve that is to create a widget that renders nothing:

class NoInput(forms.Widget):
    input_type = "hidden"
    template_name = ""

    def render(self, name, value, attrs=None, renderer=None):
        return ""

class YourForm(forms.Form):
    control = forms.CharField(required=False, widget=NoInput)

This way you can still use {{form}} in your template.

Answer 3

The best way to do what you want to do here is write a custom widget to display the file field differently. However, the best way to answer the exact question you asked, without having to render the form manually in the templates, is to overwrite the as_table method of the form:

class InsertCoupon(forms.ModelForm):

    def as_table(self, *args, **kwargs):
        del self.fields['image']
        return super(InsertCoupon, self).as_table(*args, **kwargs)

Answer 4

So a better way to exclude just a single field in the template might be:

<table>
{% for field in form %}
    {% if field != "image" %}
    <tr><td>{{ field }}</td></tr>
    {% endif %}
{% endfor %}
</table>

Beats manually writing out all the fields.