Model ImageField and save in a specific path - django

Question

This is my model:
def group_based_upload_to(instance, filename):
return "media/image/lavorazione/{}".format(instance.prestazione.id,)
class ImmaginiLavorazioni(models.Model):
immagine = models.ImageField(upload_to=group_based_upload_to)
prestazione = models.ForeignKey(Listino, on_delete=models.SET_NULL, null=True,
blank=True, default=None, related_name='Listino3')
and my form:
class ImmagineUploadForm(forms.ModelForm):
class Meta:
model = ImmaginiLavorazioni
exclude = ('preventivo', )
I need a view to save an image in a specific path.
The name of path must be the pk of foreign key.
How can I do that?

Answer 1

I use a model of how a blog post works. You can adjust the example on your needs. You should try and avoid saving the location path from a view.
On your models.py:
# Create your models here.
def upload_location(instance, filename):
filebase, extension = filename.split(".")
return "%s/%s" % (instance.id, filename)
class Post(models.Model):
title = models.CharField(max_length=120)
slug = models.SlugField(unique=True)
image = models.ImageField(upload_to=upload_location,
null=True,
blank=True,
height_field="height_field",
width_field="width_field")
height_field = models.IntegerField(default=0)
width_field = models.IntegerField(default=0)
content = HTMLField()
formfield_overrides = {
models.TextField: {'widget': AdminPagedownWidget },
}
updated = models.DateTimeField(auto_now=True, auto_now_add=False)
timestamp = models.DateTimeField(auto_now=False, auto_now_add=True)
On your forms.py:
from django import forms
from pagedown.widgets import PagedownWidget
from pagedown.widgets import AdminPagedownWidget
from .models import Post
class PostForm(forms.ModelForm):
content = forms.CharField(widget=PagedownWidget(show_preview=False))
class Meta:
model = Post
fields = [
"title",
"content",
"image"
]
class PostModelForm(forms.ModelForm):
content = forms.CharField(widget=AdminPagedownWidget())
class Meta:
model = Post
fields = '__all__'
And on your settings.py:
MEDIA_URL = "/media/"
MEDIA_ROOT = os.path.join(os.path.dirname(BASE_DIR), "media_cdn")
Here is your view.py:
# Create your views here.
def post_create(request):
if not request.user.is_staff or not request.user.is_superuser:
raise Http404
form = PostForm(request.POST or None, request.FILES or None)
if form.is_valid():
instance = form.save(commit=False)
instance.save()
messages.success(request, "Succefully Created")
return HttpResponseRedirect(instance.get_absolute_url())
context = {
"form": form,
}
return render(request, "post_form.html", context)

Answer 2

The next code has an example without form but you can modify based on your needs.
Below the settings.py
# Build paths inside the project like this: os.path.join(BASE_DIR, ...)
BASE_DIR = os.path.dirname(os.path.realpath(__file__))
MEDIA_ROOT = os.path.join(BASE_DIR, '../media')
Firstable, I defined the path of the image in the helpers.py
from django.conf import settings
from datetime import datetime
def upload_to_image_post(self, filename):
"""
Stores the image in a specific path regards to date
and changes the name of the image with for the name of the post
"""
current_date = datetime.now()
return '%s/posts/main/{year}/{month}/{day}/%s'.format(
year=current_date.strftime('%Y'), month=current_date.strftime('%m'),
day=current_date.strftime('%d')) % (settings.MEDIA_ROOT, filename)
You could define the image's name with some code like this one. But you have to regard you should have the pk to replace the name of the image
ext = filename.split('.')[-1]
name = self.pk
filename = '%s.%s' % (name, ext)
So, I called the def in my models.py, specifically in the image's field
from django.db import models
from django.utils.text import slugify
from .helpers import upload_to_image_post
class Post(models.Model):
"""
Store a simple Post entry.
"""
title = models.CharField('Title', max_length=200, help_text='Title of the post')
body = models.TextField('Body', help_text='Enter the description of the post')
slug = models.SlugField('Slug', max_length=200, db_index=True, unique=True, help_text='Title in format of URL')
image_post = models.ImageField('Image', max_length=80, blank=True, upload_to=upload_to_image_post, help_text='Main image of the post')
class Meta:
verbose_name = 'Post'
verbose_name_plural = 'Posts'
I hope this helped you

Answer 3

Now this is my models:
def group_based_upload_to(instance, immagine):
return "media/preventivo/{pk}/{image}".format(pk=instance.preventivo.id, image=immagine)
class ImmaginiLavorazioni(models.Model):
immagine = models.ImageField(upload_to=group_based_upload_to)
preventivo = models.ForeignKey(Preventivo, on_delete=models.SET_NULL, null=True, blank=True,
default=None, related_name='Listino3')
And this is my view:
def upload_immagine(request, pk):
member = get_object_or_404(Preventivo, pk=pk)
if request.method == 'POST':
form = ImmagineUploadForm(request.POST or None, request.FILES or None)
if form.is_valid():
newimmagine = form.save(commit=False)
newimmagine.preventivo_id = member.pk
newimmagine.save()
return redirect('preventivo_new2', pk=member.pk)
else:
form = ImmagineUploadForm(request.POST or None, request.FILES or None)
return render(request, 'FBIsystem/upload_immagine.html', {'member': member, 'form': form})
It save in db the file path and the foreign key like:
immagine= media/preventivo/6/image.jpg
preventivo_id = 6
but not create a folder and not save the uploaded image...