How to integrate a form into a Detail View? - django

Question

I would like to do:
I am trying to create a form input on a detail view that will update a particular data column ('status') of the detailed model instance. Here is a picture of what I have in mind:
The selector would display the current status and the user could change it and update from the detail view without having to access the UpdateView.
my idea here would be to have this happen:
1. On submit, get the new user entered value.
2. get the model instance of the currently detailed class
3. assign the model instance attribute as the user entered value
4. save the model instance
I've tried: I don't know if this is the best way to do this but i've been trying to create an AJAX call, mostly by looking for examples online.
Results: Terminal shows Post on submit: "[19/Nov/2019 17:50:33] "POST /task/edit/4 HTTP/1.1" 200 41256". However, the data is not saved to the db. On refresh, the selector returns to previously saved status.
The console shows: "script is connected", and "Update Status" with no errors. On submit, the alert displays success message: "127.0.0.1:8000 says status updated".
Task_detail.html
<div class="deliv-box edit">
<form id="status-update-form" method='POST' action='{% url "task_edit" task.pk %}'>
{% csrf_token %}
{{task_form.status}}
<input id="status-update-btn" type="submit" value="Update Status" />
</form>
</div>
...
<script type="text/javascript">
var frm = $('#status-update-form');
frm.submit(function () {
console.log("script is connected")
console.log($('#status-update-btn').val())
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
$("#deliv-box edit").html(data);
alert("status updated");
},
error: function(data) {
alert("error");
}
});
return false;
});
</script>
forms.py
class TaskForm(forms.ModelForm):
class Meta:
model = Task
fields = "__all__"
views.py
class TaskDetail(ModelFormMixin, DetailView):
template_name='task_detail.html'
model = Task
form_class = TaskForm
def get_context_data(self, **kwargs):
context = super(TaskDetail, self).get_context_data(**kwargs)
context['task_form'] = self.get_form
return context
def update(request):
if request.method=='POST':
task_id = request.POST.get('id')
task = Task.objects.get(pk = task_id)
status_obj = request.POST.get('status')
task.status = status_obj
task.save()
return JsonResponse({'status':'updated...'})
else:
return JsonResponse({'status':'not updated'})
thank you.
A solution:
In the unlikely event that someone stumbles across this question and who is, like me, just trying to figure it out all by themselves, here is what I've learned about how this works: When a user wants to update a form, Django pre-populates the form with the existing data related to that instance. A user can then alter the data and re-submit the form.
Here, I was attempting to alter just one field of the exiting instance, but as I was only calling that one field, Django was assuming not, as I had hoped, that the other fields would remain the same, but that I intended the other fields to be submitted as blank. Where the fields are required one cannot return that field as blank. Therefore, Django was not able to validate the form and so the form did not get updated.
A solution that works is to call all the fields as hidden and show just the one you want to alter. This way Django can return the unaltered data and validate the form, and you get an update button on your detail view:
<form method="POST">
{% csrf_token %}
<h4> STATUS: </h4>
{% for field in form %}
{{ field.as_hidden }}
{% endfor %}
{{form.status}}
<button type="submit" class="btn btn-success">submit</button>
</form>

Answer 1

You are overriding the method update which does not exist, so it is never called.
You need to subclass UpdateView instead of the DetailView and the mixin.
class TaskUpdateView(UpdateView):
template_name='task_detail.html'
model = Task
form_class = TaskForm
# you can use the line below instead of defining form_class to generate a model form automatically
# fields = ('status', )
def form_valid(self, form):
post = form.save(commit=False)
# do anything here before you commit the save
post.save()
# or instead of two lines above, just do post = form.save()
return JsonResponse({'status':'updated...'})
Here is how you would add readonly (disabled) fields to your form:
class TaskForm(forms.ModelForm):
# override the default form field definitions for readonly fields
other_field = forms.CharField(disabled=True)
another_field = forms.IntegerField(disabled=True)
class Meta:
model = Task
fields = ("status", "other_field", "another_field")
# you could also just do:
# fields = '__all__'