Question

I’m new to Django but I seem to have nearly identical code working on another site. I can update a record in the Django shell, but in view.py the same code insists on INSERTing a new record when I run this form.

• I have a “DisciplineEvent” object in the model. I let Django create the “id” field for the primary key. I can see the “id” int column has been created with auto_increment in MySQL
• The form DisciplineEventEntryForm is created from the “DisciplineEvent” model object.
• To edit a record, the entry form is populated and the pk is put in a hidden field named “id”, which appears to be submitted along with the POST data.

So the relevant part of the view.py is this:

if request.method == 'POST':
    incidentId = request.POST['id']
    editedEvent = DisciplineEvent.objects.get(pk=int(incidentId))
    form = DisciplineEventEntryForm(request.POST, instance=editedEvent)
    form.save()
    variables = Context({
        'account': account,
        'date': request.POST['event_date'],
        'description': request.POST['incident_description'],
        'incident_id':incidentId,
         })
     template = get_template('disciplineform_confirm_entry.html')
     output = template.render(variables)
     response = HttpResponse(output)
     return response

I thought this would pull the record in question, save the new form data into it, and UPDATE the record. Instead it creates a new record with all the data and an incremented primary key.

Answer 1

What you are trying to do is unconventional and a possible security hole.

You should not get the instance of the object from the hidden id key you populated in the form. Users can easily change this one and get your code to overwrite some other model instance that they may not even have permission for.

The standard way to do it is to obtain the object based on the url.

def view_function(request,id):
    object_to_edit = get_object_or_404(Model,id=id) #Or slug=slug
    form = ModelForm(data = request.POST or None, instance=object_to_edit)
    if form.is_valid():
        form.save()
        redirect()
    return render_to_response('template_name',{},RequestContext(request))

Hope it helps!