django filter != choicefield code example

Example: django choice field

First I recommend you as @ChrisHuang-Leaver suggested to define a new file with all the choices you need it there, like choices.py:

STATUS_CHOICES = (
    (1, _("Not relevant")),
    (2, _("Review")),
    (3, _("Maybe relevant")),
    (4, _("Relevant")),
    (5, _("Leading candidate"))
)
RELEVANCE_CHOICES = (
    (1, _("Unread")),
    (2, _("Read"))
)
Now you need to import them on the models, so the code is easy to understand like this(models.py):

from myApp.choices import * 

class Profile(models.Model):
    user = models.OneToOneField(User)    
    status = models.IntegerField(choices=STATUS_CHOICES, default=1)   
    relevance = models.IntegerField(choices=RELEVANCE_CHOICES, default=1)
And you have to import the choices in the forms.py too:

forms.py:

from myApp.choices import * 

class CViewerForm(forms.Form):

    status = forms.ChoiceField(choices = STATUS_CHOICES, label="", initial='', widget=forms.Select(), required=True)
    relevance = forms.ChoiceField(choices = RELEVANCE_CHOICES, required=True)
Anyway you have an issue with your template, because you're not using any {{form.field}}, you generate a table but there is no inputs only hidden_fields.

When the user is staff you should generate as many input fields as users you can manage. I think django form is not the best solution for your situation.

I think it will be better for you to use html form, so you can generate as many inputs using the boucle: {% for user in users_list %} and you generate input with an ID related to the user, and you can manage all of them in the view