Can someone help me with fixing Django ModelForm?
This particular code can add new item to database as expected, but when I'm trying to edit db record - It just add new record, instead of updating old. I'm quite new in Django framework.
views.py:
def manage(request, item_id = None):
t = get_object_or_404(Hardware, id=item_id) if item_id else None
form = Manage(request.POST or None, instance=t)
if t:
if form.is_valid():
#form.save()
hostname = form.cleaned_data['hostname']
cpu = form.cleaned_data['cpu']
os = form.cleaned_data['os']
ram = form.cleaned_data['ram_total']
storage = form.cleaned_data['storage']
hostdata = Hardware(
hostname=hostname,
cpu=cpu,
ram_total=ram,
os=os,
storage=storage,
lock_state=t.lock_state, # because in edit operation we shouldn't change it.
lock_date=t.lock_date, # because in edit operation we shouldn't change it.
locked_by=t.locked_by) # because in edit operation we shouldn't change it.
hostdata.save()
return HttpResponseRedirect(reverse('main:index'))
elif not t:
if form.is_valid():
hostname = form.cleaned_data['hostname']
cpu = form.cleaned_data['cpu']
os = form.cleaned_data['os']
ram = form.cleaned_data['ram_total']
storage = form.cleaned_data['storage']
current_user = request.user
user = User.objects.get(id=current_user.id)
hostdata = Hardware(
hostname=hostname,
cpu=cpu,
ram_total=ram,
os=os,
storage=storage,
lock_state=0,
lock_date=datetime.datetime.now(),
locked_by=user)
hostdata.save()
return HttpResponseRedirect(reverse('main:index'))
return render(request, 'hardware/edit.html', {'form': form})
models.py:
class Hardware(models.Model):
hostname = models.CharField(max_length=255, default=None)
os = models.CharField(max_length=255, default=None)
cpu = models.CharField(max_length=255, default=None)
ram_total = models.CharField(max_length=255, default=None)
storage = models.CharField(max_length=255, default=None)
lock_state = models.BooleanField(default=0)
locked_by = models.ForeignKey(User)
lock_date = models.DateTimeField(default=None)
alive = models.BooleanField(default=0)
class Meta:
db_table = "hardware"
def __str__(self):
return self.hostname
forms.py:
class Manage(forms.ModelForm):
class Meta:
model = Hardware
fields = ['hostname', 'os', 'cpu', 'ram_total', 'storage']
urls.py:
url(r'^manage/new/$', views.manage, name='add'),
url(r'^manage/edit/(?P<item_id>[0-9]+)/$', views.manage, name='edit')
template:
<form action="" method="post">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Save!" />
</form>
You already retrieved the instance t in the first line of your view. The code below will always create a new instance (unless you specify the pk parameter):
hostdata = Hardware(...)
hostdata.save()
Simply do this instead:
if t:
if form.is_valid():
t.hostname = form.cleaned_data['hostname']
t.cpu = form.cleaned_data['cpu']
....
t.save()
However, you really should rely on the save method provided by the ModelForm as the other answers suggested. Here's an example:
def manage(request, item_id=None):
t = get_object_or_404(Hardware, id=item_id) if item_id else None
# if t is None, a new object will be created in form.save()
# if t is an instance of Hardware, t will be updated in form.save()
form = Manage(request.POST, instance=t)
if form.is_valid():
form.save()
return HttpResponseRedirect(reverse('main:index')
return render(request, 'hardware/edit.html', {'form': form})
You also specified fields in your form:
fields = ['hostname', 'os', 'cpu', 'ram_total', 'storage']
These are the fields which will be set or updated when you call form.save().
I think something like this - using update_fields - should work:
def manage(request, item_id = None):
t = get_object_or_404(Hardware, id=item_id)
form = Manage(request.POST or None, instance=t)
if t:
if form.is_valid():
#form.save()
t.hostname = form.cleaned_data['hostname']
t.cpu = form.cleaned_data['cpu']
t.os = form.cleaned_data['os']
t.ram = form.cleaned_data['ram_total']
t.storage = form.cleaned_data['storage']
t.save(update_fields=['hostname', 'cpu', 'os','ram','storage'])
return HttpResponseRedirect(reverse('main:index'))
........
Try Class Based View, which in it's simplest looks like:
from django.views import generic
class HardwareEditView(generic.UpdateView):
template_name = "hardware.html"
form_class = Manage
You will have to add get_absolute_url to the model.
Generic class based views are exactly for this standard create/update/view common tasks.
Related
I'm running some tests for my app 'ads', but when I try to test the CreateView it fails with the following message:
AssertionError: 'just a test' != 'New title'
Here's the test:
class AdTests(TestCase):
def setUp(self):
self.user = get_user_model().objects.create_user(
username='test_user',
email='test#email.com',
password='secret'
)
self.ad = Ad.objects.create(
title='just a test',
text='Ehy',
owner=self.user
)
def test_ad_create_view(self):
response = self.client.post(reverse('ads:ad_create'), {
'title': 'New title',
'text': 'New text',
'owner': self.user.id,
})
self.assertEqual(response.status_code, 302)
self.assertEqual(Ad.objects.last().title, 'New title')
self.assertEqual(Ad.objects.last().text, 'New text')
So it could be that the test fails in creating a new ad, and then it compares the fields with the first ad in the setUp method.
I upload the rest of the code if it can help:
urls.py
from django.urls import path, reverse_lazy
from . import views
app_name='ads'
urlpatterns = [
path('', views.AdListView.as_view(), name='all'),
path('ad/<int:pk>', views.AdDetailView.as_view(), name='ad_detail'),
path('ad/create',
views.AdCreateView.as_view(success_url=reverse_lazy('ads:all')), name='ad_create'),
...
]
models.py
class Ad(models.Model) :
title = models.CharField(
max_length=200,
validators=[MinLengthValidator(2, "Title must be greater than 2 characters")]
)
price = models.DecimalField(max_digits=7, decimal_places=2, null=True)
text = models.TextField()
"""We use AUTH_USER_MODEL (which has a default value if it is not specified in settings.py) to create a Foreign Key relationship between the Ad model
and a django built-in User model"""
owner = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete=models.CASCADE)
comments = models.ManyToManyField(settings.AUTH_USER_MODEL, through='Comment', related_name='comments_owned')
picture = models.BinaryField(null=True, editable=True)
tags = TaggableManager(blank=True)
content_type = models.CharField(max_length=256, null=True, help_text='The MIMEType of the file')
favorites = models.ManyToManyField(settings.AUTH_USER_MODEL, through='Fav', related_name='favorite_ads')
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
views.py
class AdCreateView(LoginRequiredMixin, View):
template_name = 'ads/ad_form.html'
success_url = reverse_lazy('ads:all')
def get(self, request, pk=None):
form = CreateForm()
ctx = {'form': form}
return render(request, self.template_name, ctx)
# Pull data
def post(self, request, pk=None):
form = CreateForm(request.POST, request.FILES or None)
if not form.is_valid():
ctx = {'form': form}
return render(request, self.template_name, ctx)
pic = form.save(commit=False)
pic.owner = self.request.user
pic.save()
form.save_m2m()
return redirect(self.success_url)
forms.py (the view uses it especially to check and save the image)
class CreateForm(forms.ModelForm):
max_upload_limit = 2 * 1024 * 1024
max_upload_limit_text = naturalsize(max_upload_limit)
# Call this 'picture' so it gets copied from the form to the in-memory model
# It will not be the "bytes", it will be the "InMemoryUploadedFile"
# because we need to pull out things like content_type
picture = forms.FileField(required=False, label='File to Upload <= ' + max_upload_limit_text)
upload_field_name = 'picture'
class Meta:
model = Ad
fields = ['title', 'text', 'price', 'picture', 'tags']
# Check if the size of the picture is less than the one specified (see above).
def clean(self):
cleaned_data = super().clean()
pic = cleaned_data.get('picture')
if pic is None:
return
if len(pic) > self.max_upload_limit:
self.add_error('picture', "File must be < " + self.max_upload_limit_text + " bytes")
# Convert uploaded File object to a picture
def save(self, commit=True):
instance = super(CreateForm, self).save(commit=False)
# We only need to adjust picture if it is a freshly uploaded file
f = instance.picture # Make a copy
if isinstance(f, InMemoryUploadedFile): # Extract data from the form to the model
bytearr = f.read()
instance.content_type = f.content_type
instance.picture = bytearr # Overwrite with the actual image data
if commit:
instance.save()
self.save_m2m()
return instance
I hope it is useful, thanks in advance!
According to Django Doc
Create View:
A view that displays a form for creating an object, redisplaying the form with validation errors (if there are any) and saving the object.
This is not a Valid way to do create View based on my experience. Check the doc Doc here.
if i understand what you are talking about You want to submit Ad Model using Create View, if you want to submit it in form You can something like this:
from django.views.generic import CreateView
from django.contrib.auth.mixins import LoginRequiredMixin
from django.urls import reverse_lazy,
class PostCreativeView(LoginRequiredMixin, CreateView):
model = #Your Model
fields = [#Fields of the model You want to submit]
template_name = #html template you want to submit the form
success_url = reverse_lazy(#url for redirected user when the form is submitted)
def form_valid(self, form):
form.instance.user = self.request.user
return super (PostCreativeView, self).form_valid(form)
in the form template you can add:
<form method="post">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Save">
</form>
for styling you can follow this: Answer
I'm working on a project "Beauty Parlour Management System" and I got this error (Cannot assign "'7'": "Appointment.your_service" must be a "Service" instance.) anyone here can help me, please.
When I am filling a book appointment form then I got this error.
models.py
class Service(models.Model):
name = models.CharField(max_length=50)
price = models.IntegerField(default=0)
image = models.ImageField(upload_to='uploads/productImg')
class Appointment(models.Model):
your_name = models.CharField(max_length=100)
your_phone = models.CharField(max_length=10)
your_email = models.EmailField(max_length=200)
your_service = models.ForeignKey('Service', on_delete=models.CASCADE, default=1)
your_date = models.DateField()
views.py
def appointments(request):
if request.method == 'GET':
return render(request, 'core/bookappointment.html')
else:
your_name = request.POST.get('your-name')
your_phone = request.POST.get('your-phone')
your_email = request.POST.get('your-email')
your_service = request.POST.get('your-service')
your_date = request.POST.get('your-date')
details = Appointment(
your_name = your_name,
your_phone = your_phone,
your_email = your_email,
your_service = your_service,
your_date = your_date)
details.save()
return render(request, 'core/appointments.html')
You create this by assigining the method to your_service_id field, if you work with your_service, it should be a Service object:
details = Appointment.objects.create(
your_name=your_name,
your_phone=your_phone,
your_email=your_email,
your_service_id=your_service,
your_date=your_date
)
That being said, it is usually better to validate, clean, and save the data with a ModelForm, not manually.
Note: In case of a successful POST request, you should make a redirect
[Django-doc]
to implement the Post/Redirect/Get pattern [wiki].
This avoids that you make the same POST request when the user refreshes the
browser.
def appointments(request,pk):
record = get_object_or_404(Service,pk=pk)
if request.method == 'POST':
form = appointmentsForm(request.POST,request.FILES)
if form.is_valid():
appointment= form.save(commit=False)
appointment.your_service = record
appointment.save()
return render(request, 'core/bookappointment.html')
else:
return render(request, 'core/appointments.html')
Hi I was just learning django and trying to create a model form with manaytomany relationship between item and order
below is my code snippet.
models.py
class Item(models.Model):
name = models.CharField(max_length=25,default="",primary_key=True)
weight = models.FloatField()
def __str__(self):
return self.name
class Order(models.Model):
customername = models.CharField(max_length=25,default="")
item = models.ManyToManyField(Item,default="")
metal = models.ForeignKey(Material,on_delete=models.CASCADE)
price = models.IntegerField()
place = models.CharField(max_length=25)
orderdate = models.DateTimeField(auto_now_add=True)
def __str__(self):
return self.customername
forms.py:
from django import forms
from .models import Order, Material, Item
class ItemForm(forms.ModelForm):
class Meta:
model = Item
fields = '__all__'
class OrderForm(forms.ModelForm):
class Meta:
model = Order
fields = '__all__'
views.py:
def ItemSaveorUpdate(request):
if request.method == 'POST':
form = ItemForm(request.POST)
if form.is_valid():
form.save()
messages.add_message(request,messages.SUCCESS,"Movie rating submitted succesfully")
else:
form = ItemForm()
return render(request,'addItem.html',{"form":form})
def OrderSaveorUpdate(request):
if request.method == 'POST':
form = OrderForm(request.POST)
if form.is_valid():
form.save()
messages.add_message(request,messages.SUCCESS,"Order added succesfully")
return redirect('material')
elif request.method == 'PUT':
item = Item.objects.get(pk=id)
form = OrderForm(instance=item)
form.save()
else:
#item = Item.objects.get(pk=id)
form = OrderForm()
return render(request,'addOrder.html',{"form":form})
Template additem.html:
<form action="{% url 'item' %}" request="post">
{% csrf_token %}
{{form.as_p }}
<input type="submit" value="add">
</form>
urls.py:
urlpatterns = [
path('item/', views.ItemSaveorUpdate, name="item"),
path('material/', views.MaterialSaveorUpdate, name="material"),
path('order/', views.OrderSaveorUpdate, name="order"),
I am trying to select multiple items while creating an order and after clicking add it is giving the error IntegrityError at /collection/order/
FOREIGN KEY constraint failed. Important to mention that the order is getting added which I can see in admin panel but the items are not getting selected.
Please help to avoid this issue and let me know what I missed here.
Thank you.
The issue solved by removing default for item fields in Order class in models.py
item = models.ManyToManyField(Item,default="")
after removing
item = models.ManyToManyField(Item)
Its is working fine now.
Thank you.
So I'm building a basic Q&A site-- Each topic has a series of questions associated with it, and each question has multiple answers associated with it.
I'm creating the user input for questions and they have to associated with a topic. This is the questions model
#models.py
class Question(models.Model):
movie = models.ForeignKey(Movie, blank=True, null=True)
question_text = models.CharField(max_length = 1000)
question_detail = models.CharField(max_length = 5000, blank = True, null = True)
q_pub_date = models.DateTimeField(auto_now_add = True)
q_author = models.ForeignKey(User)
class QuestionForm(ModelForm):
def save(self, user = None, force_insert = False, force_update = False, commit = True):
q = super(QuestionForm, self).save(commit = False)
q.q_author = user
if commit:
q.save()
return q
class Meta:
model = Question
exclude = ('movie', 'q_author', 'q_pub_date')
This is the URL conf
#urls.py
url(r'^(?P<movie_id>\d+)/add_question/$', 'add_question'),
Now here is the view
#views.py
def add_question(request, movie_id):
if request.method == "POST":
form = QuestionForm(request.POST, request.FILES)
#QuestionForm.movie = Movie.objects.get(pk = movie_id)
if form.is_valid():
form.save(user = request.user)
return HttpResponseRedirect("/home/")
else:
form = QuestionForm()
return render_to_response("qanda/add_question.html", {'form': form}, context_instance = RequestContext(request))
This is the HTML code
#add_question.html
<h1> Add Question: {{ user.username }}</h1>
<form action = "" method = "post">{% csrf_token %}
{{ form.as_p }}
<input type = "submit" value = "Ask" />
<input type = "hidden" name = "next" value = "{{ next|escape }}" />
</form>
In the view, the commented out line is what I added to the view to try and auto save the model. When adding a question, the URL has the ID of the movie it is associated with, and my thought is to take that ID and then plug it into the ForeignKey to identify which movie is associated with the question. However, when I use my code, it changes all of the Questions' movie associations to the current movie instead of just changing that specific question's movie association. Without the code, it doesn't associate a Movie with the Question at all. How do I fix this?
Use this:
#views.py
def add_question(request, movie_id):
if request.method == "POST":
form = QuestionForm(request.POST, request.FILES)
if form.is_valid():
question = form.save(user = request.user)
question.movie = Movie.objects.get(pk = movie_id)
question.save()
return HttpResponseRedirect("/home/")
else:
form = QuestionForm()
return render_to_response("qanda/add_question.html", {'form': form}, context_instance = RequestContext(request)
For question asked in comment
You should avoid using absolute URLs in views or templates. Consider a scenario, where you decide to change home URL from /home/ to /myhome/. You will have to edit it where ever you have used them. It is always better to name the urls (docs):
# URL Conf
url(r'^home/$', 'home_view', name="home_url"),
url(r'^(?P<movie_id>\d+)/add_question/$', 'add_question', name="add_question_url"),
url(r'^home/(?P<movie_id>\d+)/$', 'movie_view', name="movie_url"),
The name argument act as an unique identifier to your actual URLs
Now in you views:
from django.core.urlresolvers import reverse
def some_view(request):
...
return HttpResponseRedirect(reverse('home_url'))
Now what ever change you make to the URL (say /home/ to /myhome/ makes no effect to the view as long as the name argument has the same value in the URL conf.
If you wish to pass parameters (like movie_id in your case)
def some_view(request, movie_id):
...
return HttpResponseRedirect(reverse('movie_url', kwargs={'movie_id':movie_id}))
The same concept should be used in templates to avoid hard-coding URLS in templates. Please read this for more details
I'm trying to make a view where the user can edit DB records through a form in a template. I've searched a lot of web pages (and Django docs as well) where they teach how to make these views, but they always use the "id" that Django generates for each Model. In this particular Model, I have to use an AutoField to override the "id". Is there a way to use this AutoField as an "id" of the record with Django?
Here's my complete model:
class T031003 (models.Model):
C003IDCD = AutoField(primary_key=True)
C003INST = models.IntegerField(unique=True) #usar AutoSlug
C003TPCD = models.CharField(max_length=1)
C003CHCD = models.CharField(max_length=14)
C003MTR = models.CharField(max_length=30, blank=True, null=True)
C003CTCD = models.CharField(max_length=3)
C003RZSC = models.CharField(max_length=60, blank=True, null=True)
C003EML = models.EmailField(max_length = 254, blank=True, null=True)
C003LOGA = models.CharField(max_length=20)
C003LOGB = models.DateTimeField()
C003LOGD = models.CharField(max_length=15, blank=True, null=True)
C003LOGF = models.CharField(max_length=20, blank=True, null=True)
def __unicode__(self):
return '%s' % self.C003MTR
class T031003Form(ModelForm):
class Meta:
model = T031003
ordering = ["-C003MTR"]
exclude = ('C003LOGA','C003LOGB','C003LOGD','C003LOGE','C003LOGF')
And here's the view I tried to do, but it gives me the error "No T031003 matches the given query." and it's right, since there is no "id" in the table:
def t031003form_edit(request, id=None):
pin = get_object_or_404(T031003, pk=id)
form = T031003Form(request.POST or None, instance=pin)
if request.method == 'POST':
if form.is_valid():
form = form.save(False)
form.C003LOGA = request.user
form.C003LOGB = datetime.date.today()
form.C003LOGD = request.META['REMOTE_ADDR']
form.C003LOGF = request.META['USERDOMAIN']
form.save()
form = T031003Form()
else:
return HttpResponseRedirect('/erro/')
return render_to_response('T031003Form_edit.html', {'form': form,}, context_instance=RequestContext(request))
Any help would be very appreciated!
If a model has an AutoField — an auto-incrementing primary key — then that auto-incremented value will be calculated and saved as an attribute on your object the first time you call save():
>>> b2 = Blog(name='Cheddar Talk', tagline='Thoughts on cheese.')
>>> b2.id # Returns None, because b doesn't have an ID yet.
>>> b2.save()
>>> b2.id # Returns the ID of your new object.
There's no way to tell what the value of an ID will be before you call save(), because that value is calculated by your database, not by Django.
ref : https://docs.djangoproject.com/en/dev/ref/models/instances/?from=olddocs
Well, thanks to the help from a close friend, I could do the trick using formsets. Here's the view:
def t031002form_edit(request, id_auto):
j = get_object_or_404(T031002, pk=id_auto)
T031003FormSet = modelformset_factory(T031002, can_delete=True, max_num=1)
if request.method == 'POST':
form = T031002FormSet(request.POST or None, request.FILES or None, queryset=T031002.objects.filter(pk=id_auto))
if form.is_valid():
instance = form.save(commit=False)
form.C003LOGA = request.user
form.C003LOGB = datetime.date.today()
form.C003LOGD = request.META['REMOTE_ADDR']
form.C003LOGF = request.META['USERDOMAIN']
for reform in instance:
reform.save()
else:
return HttpResponseRedirect('/erro/')
else:
form = T031002FormSet(queryset=T031002.objects.filter(pk=id_auto))
return render_to_response(('T031002Form_edit.html'), {'form': form,}, context_instance=RequestContext(request))
So, with formsets, you can work nicely and with no worries. Hope it helps others with this same questioning.