如何在 Django 中更改模型的状态

Question

I'm building a small to-do list project using Django, the to-do list has tasks divided into to-do, in-progress and done, I want to move to-do tasks from todo list to in- progress list by changing its status, so I gave the to-do model 3 status, and below in views.py I specified a method to change the status, but it seems not work, I'm not sure how to set the status. Any help would be appreciated.

models.py

'''

class Todo(models.Model):
status_option = (
    ('to_do', 'to_do'),
    ('in_progress', 'in_progress'),
    ('done', 'done'),
)
status = models.CharField(max_length=20, choices=status_option, default='to_do')
project = models.ForeignKey(Project, on_delete=models.CASCADE)
name = models.CharField(max_length=20)
create_date = models.DateTimeField(auto_now_add=True)
start_date = models.DateTimeField(default=datetime.datetime.now)
due_date = models.DateTimeField(default=datetime.datetime.now)
details = models.TextField()

def __str__(self):
    return self.status

'''

views.py

'''

def add_to_progress(request, todo_id, project_id):
    todo = Todo.objects.get(id=todo_id)
    project = Project.objects.get(id=project_id)

    if request.method != 'POST':
        form = dragTodoForm()
    else:
        form = dragTodoForm(request.POST)
        Todo.objects.filter(id=todo_id).update(status='in_progress')
    context = {'form', form, 'todo', todo, 'project', project}
    return render(request, 'todo_lists/new_progress.html', context)


def add_to_done(request, todo_id, project_id):
    todo = Todo.objects.get(id=todo_id)
    project = Project.objects.get(id=project_id)

    if request.method != 'POST':
        form = dragTodoForm()
    else:
        form = dragTodoForm(request.POST)
        Todo.objects.filter(id=todo_id).update(status='done')
    context = {'form', form, 'todo', todo, 'project', project}
    return render(request, 'todo_lists/new_done.html', context)

'''

views.py

'''

def drag(request, project_id):
project = Project.objects.get(id=project_id)
todo = Todo.objects.filter(status='to_do')
progress = Todo.objects.filter(status='in_progress')
done = Todo.objects.filter(status='done')
context = {'todo', todo, 'progress', progress, 'done', done, 'project', 
project}
return render(request, 'todo_lists/project.html', context)

'''

Answer 1

I solved the problem using below strategy:
models.py

''' class Todo(models.Model):

status_option = (
    ('to_do', 'to_do'),
    ('in_progress', 'in_progress'),
    ('done', 'done'),
)
status = models.CharField(max_length=20, choices=status_option, default='to_do')
# todo_list's content
team = models.ForeignKey('Team', on_delete=models.CASCADE)
project = models.ForeignKey(Project, on_delete=models.CASCADE)
name = models.CharField(max_length=20)
create_date = models.DateTimeField(auto_now_add=True)
start_date = models.DateTimeField(default=datetime.datetime.now)
due_date = models.DateTimeField(default=datetime.datetime.now)

project_code = models.CharField(max_length=20)
details = models.TextField()

def __str__(self):
    return self.status
    # return self.team['team'].queryset

def update_status(self):
    if self.status == 'to_do':
        self.status = 'in_progress'
    elif self.status == 'in_progress':
        self.status = 'done'
    self.save()

'''

views.py

'''

def progress(request, pk):
    to = get_object_or_404(Todo, pk=pk)
    to.update_status()
    return redirect(reverse('todo_lists:project', kwargs={'project_id': 
to.project.pk}))

'''