If I have a Django model with a FileField e.g.,
class Probe(models.Model):
name = models.CharField(max_length=200, unique=True)
nanoz_file = models.FileField(upload_to='nanoz_file', blank=True)
is there a way to prevent an uploaded file from being overwritten if a user uploads a new file in the Admin interface?
Also if I do keep the old files around is there a way I can relate the previous files back to the model instance?
I.e., I'd like to be able to list all files uploaded to the nanoz_file field for a given model instance.
Django never overwrites an uploaded file. If you upload 'foo.png' twice, the second will be 'foo_1.png' - I just tested this but don't take my word for it: try it too !
All you have to do (or let django-reversion do) is keep track of the previous file names.
You can use this structure:
class File(models.Model):
name = models.CharField()
file = models.FileField(upload_to='files_storage/')
belongs = models.ForeignKey('self')
creation = models.DateTimeField(auto_now_add=True)
Then in the view you can use something like:
def edit_file(request, ...):
# Get the file model instance
file_model = ... # Code to get the instance
# Create a new instance of the model with the old file path
old_file = File(name='file1-v2', file=file_model.file, belongs=file_model)
old_file.save()
# Update the file_model with the new file data
Hope this helps!
Just as jpic said, you could try django-reversion, or
track names of past files in sorts of DB, for example in seperated table row, in customized field or in gfk field.
glob files online, as long as filename is managed.
For the second way, actually for dealing w/ all user uploads, its better to name the file by the pattern you designed instead of using the raw name (you could also store the raw name for later usage) . For your case, since the name field is unique, the field is suitable as the base for generating filename of the uploaded files, if it rarely changes:
import os.path
from django.hash_compat import sha_constructor
def upload_to(self, filename):
return 'nanoz_file/%s%s' % (
sha_constructor(self.name).hexdigest(), os.path.splitext(filename)[-1])
class Probe(models.Model):
name = models.CharField(max_length=200, unique=True)
nanoz_file = models.FileField(upload_to=upload_to, blank=True)
Then in your view, you could fetch the list of names of all files of Probe instance probe by
import glob
# be careful to operate directory securely
glob.glob(os.path.join(
os.path.dirname(probe.nanoz_file.path),
'%s*' % sha_constructor(probe.name).hexdigest()))
Imagine a model like this:
class CFile(models.Model):
filepath = models.FileField(upload_to=...)
collection = models.ForeignKey("FileCollection",null=True)
... # other attributes that are not relevant
def clean(self):
bname = os.path.basename
if self.collection:
cfiles = self.baseline.attachment_set.all()
with_same_basename = filter(lambda e: bname(e.filepath.path) == bname(self.filepath.path),cfiles)
if len(with_same_basename) > 0:
raise ValidationError("There already exists a file with the same name in this collection")
class FileCollection(models.Model):
name = models.CharField(max_length=255)
files= models.ManyToManyField("CFile")
I want to disallow the upload of a CFile if there already exists a CFile with the same basename, that's why I added the clean. The problem is:
I upload a CFile, with the name file1.png -> gets uploaded because no other files with this name exist
I upload another CFile, with the name file1.png -> I get the expected error that I already have a file with this name. So, I try to change the file, and upload a file with a different name ( file2.png ). The problem is, I stopped via pdb in the clean, and the model instance is still file1.png. I imagine this happens because of my ValidationError and django allows me to "correct my mistake". The problem is I cannot correct it if I cannot upload another file. How can I handle this?
EDIT: This happens in the admin area, sorry for forgetting to mention this before. I don't have anything custom ( besides inlines = [ FileInline ] ).
I think the clearest way is to declare another field in your model for filename and make it unique for every collection. Like this:
class CFile(models.Model):
filepath = models.FileField(upload_to=...)
collection = models.ForeignKey("FileCollection",null=True, related_name='files')
filename = models.CharField(max_length=255)
... # other attributes that are not relevant
class Meta:
unique_together = (('filename', 'collection'),)
def save(self, *args, **kwargs):
self.filename = bname(self.filepath.path)
super(CFile, self).save(args, kwargs)
having a similar model:
class Foo(models.Model):
slug = models.SlugField(unique=True)
img = ImageWithThumbnailsField(upload_to='uploads/',thumbnail={'size': (56, 34)})
It works fine but I want to add 2 more features to it:
1- It should also generate a second thumbnail sized 195x123, in addition to 56x34
2- While saving the model original image and it's two thumbnails should be renamed as by using the slug.
For instance
I am uploading 1.jpg and I name slug as "i-like-this-country2"
I should save these named versions should be saved:
1- i-like-this-country2_original.jpg
2- i-like-this-country2_middle.jpg #for 195x123
3- i-like-this-country2_small.jpg #for 56x34
First part:
Just pass it in like this: sizes=( (56,34), (195,123), )
Second part:
You can specify a function for the upload_to which Django will call, passing it an instance of the model and the original filename. With that, you can put together a function that renames the file based on the slug because Django will use whatever you return it instead. Untested code, but something like this:
def _Foo_img_name(instance, filename):
# grab the extension
left_path, extension = self.file.name.rsplit('.',1)
# change the filename then return
return 'uploads/%s.%s' % (instance.slug, extension)
class Foo(models.Model):
img = ImageWithThumbnailsField(upload_to=_Foo_img_name, ... )
I don't believe you can do is change the <filename>_56x34.jpg into anything but that.
All you have to do is to create a method in your models.py like this:
def rename_file(instance, filename):
extension = filename.split(".")[1]
rename = "rename_here"
return rename + "." + extension
Then in the class that extends models.Model
class MyImage(models.Model):
image = ImageField(upload_to=rename_file)
Don't forget to import from sorl.thumbnail import ImageField too!
Ok, I've tried about near everything and I cannot get this to work.
I have a Django model with an ImageField on it
I have code that downloads an image via HTTP (tested and works)
The image is saved directly into the 'upload_to' folder (the upload_to being the one that is set on the ImageField)
All I need to do is associate the already existing image file path with the ImageField
I've written this code about 6 different ways.
The problem I'm running into is all of the code that I'm writing results in the following behavior:
(1) Django will make a 2nd file, (2) rename the new file, adding an _ to the end of the file name, then (3) not transfer any of the data over leaving it basically an empty re-named file. What's left in the 'upload_to' path is 2 files, one that is the actual image, and one that is the name of the image,but is empty, and of course the ImageField path is set to the empty file that Django try to create.
In case that was unclear, I'll try to illustrate:
## Image generation code runs....
/Upload
generated_image.jpg 4kb
## Attempt to set the ImageField path...
/Upload
generated_image.jpg 4kb
generated_image_.jpg 0kb
ImageField.Path = /Upload/generated_image_.jpg
How can I do this without having Django try to re-store the file? What I'd really like is something to this effect...
model.ImageField.path = generated_image_path
...but of course that doesn't work.
And yes I've gone through the other questions here like this one as well as the django doc on File
UPDATE
After further testing, it only does this behavior when running under Apache on Windows Server. While running under the 'runserver' on XP it does not execute this behavior.
I am stumped.
Here is the code which runs successfully on XP...
f = open(thumb_path, 'r')
model.thumbnail = File(f)
model.save()
I have some code that fetches an image off the web and stores it in a model. The important bits are:
from django.core.files import File # you need this somewhere
import urllib
# The following actually resides in a method of my model
result = urllib.urlretrieve(image_url) # image_url is a URL to an image
# self.photo is the ImageField
self.photo.save(
os.path.basename(self.url),
File(open(result[0], 'rb'))
)
self.save()
That's a bit confusing because it's pulled out of my model and a bit out of context, but the important parts are:
The image pulled from the web is not stored in the upload_to folder, it is instead stored as a tempfile by urllib.urlretrieve() and later discarded.
The ImageField.save() method takes a filename (the os.path.basename bit) and a django.core.files.File object.
Let me know if you have questions or need clarification.
Edit: for the sake of clarity, here is the model (minus any required import statements):
class CachedImage(models.Model):
url = models.CharField(max_length=255, unique=True)
photo = models.ImageField(upload_to=photo_path, blank=True)
def cache(self):
"""Store image locally if we have a URL"""
if self.url and not self.photo:
result = urllib.urlretrieve(self.url)
self.photo.save(
os.path.basename(self.url),
File(open(result[0], 'rb'))
)
self.save()
Super easy if model hasn't been created yet:
First, copy your image file to the upload path (assumed = 'path/' in following snippet).
Second, use something like:
class Layout(models.Model):
image = models.ImageField('img', upload_to='path/')
layout = Layout()
layout.image = "path/image.png"
layout.save()
tested and working in django 1.4, it might work also for an existing model.
Just a little remark. tvon answer works but, if you're working on windows, you probably want to open() the file with 'rb'. Like this:
class CachedImage(models.Model):
url = models.CharField(max_length=255, unique=True)
photo = models.ImageField(upload_to=photo_path, blank=True)
def cache(self):
"""Store image locally if we have a URL"""
if self.url and not self.photo:
result = urllib.urlretrieve(self.url)
self.photo.save(
os.path.basename(self.url),
File(open(result[0], 'rb'))
)
self.save()
or you'll get your file truncated at the first 0x1A byte.
Ok, If all you need to do is associate the already existing image file path with the ImageField, then this solution may be helpfull:
from django.core.files.base import ContentFile
with open('/path/to/already/existing/file') as f:
data = f.read()
# obj.image is the ImageField
obj.image.save('imgfilename.jpg', ContentFile(data))
Well, if be earnest, the already existing image file will not be associated with the ImageField, but the copy of this file will be created in upload_to dir as 'imgfilename.jpg' and will be associated with the ImageField.
Here is a method that works well and allows you to convert the file to a certain format as well (to avoid "cannot write mode P as JPEG" error):
import urllib2
from django.core.files.base import ContentFile
from PIL import Image
from StringIO import StringIO
def download_image(name, image, url):
input_file = StringIO(urllib2.urlopen(url).read())
output_file = StringIO()
img = Image.open(input_file)
if img.mode != "RGB":
img = img.convert("RGB")
img.save(output_file, "JPEG")
image.save(name+".jpg", ContentFile(output_file.getvalue()), save=False)
where image is the django ImageField or your_model_instance.image
here is a usage example:
p = ProfilePhoto(user=user)
download_image(str(user.id), p.image, image_url)
p.save()
Hope this helps
What I did was to create my own storage that will just not save the file to the disk:
from django.core.files.storage import FileSystemStorage
class CustomStorage(FileSystemStorage):
def _open(self, name, mode='rb'):
return File(open(self.path(name), mode))
def _save(self, name, content):
# here, you should implement how the file is to be saved
# like on other machines or something, and return the name of the file.
# In our case, we just return the name, and disable any kind of save
return name
def get_available_name(self, name):
return name
Then, in my models, for my ImageField, I've used the new custom storage:
from custom_storage import CustomStorage
custom_store = CustomStorage()
class Image(models.Model):
thumb = models.ImageField(storage=custom_store, upload_to='/some/path')
A lot of these answers were outdated, and I spent many hours in frustration (I'm fairly new to Django & web dev in general). However, I found this excellent gist by #iambibhas: https://gist.github.com/iambibhas/5051911
import requests
from django.core.files import File
from django.core.files.temp import NamedTemporaryFile
def save_image_from_url(model, url):
r = requests.get(url)
img_temp = NamedTemporaryFile(delete=True)
img_temp.write(r.content)
img_temp.flush()
model.image.save("image.jpg", File(img_temp), save=True)
Another possible way to do that:
from django.core.files import File
with open('path_to_file', 'r') as f: # use 'rb' mode for python3
data = File(f)
model.image.save('filename', data, True)
If you want to just "set" the actual filename, without incurring the overhead of loading and re-saving the file (!!), or resorting to using a charfield (!!!), you might want to try something like this --
model_instance.myfile = model_instance.myfile.field.attr_class(model_instance, model_instance.myfile.field, 'my-filename.jpg')
This will light up your model_instance.myfile.url and all the rest of them just as if you'd actually uploaded the file.
Like #t-stone says, what we really want, is to be able to set instance.myfile.path = 'my-filename.jpg', but Django doesn't currently support that.
This is might not be the answer you are looking for. but you can use charfield to store the path of the file instead of ImageFile. In that way you can programmatically associate uploaded image to field without recreating the file.
With Django 3,
with a model such as this one:
class Item(models.Model):
name = models.CharField(max_length=255, unique=True)
photo= models.ImageField(upload_to='image_folder/', blank=True)
if the image has already been uploaded, we can directly do :
Item.objects.filter(...).update(photo='image_folder/sample_photo.png')
or
my_item = Item.objects.get(id=5)
my_item.photo='image_folder/sample_photo.png'
my_item.save()
You can try:
model.ImageField.path = os.path.join('/Upload', generated_image_path)
class tweet_photos(models.Model):
upload_path='absolute path'
image=models.ImageField(upload_to=upload_path)
image_url = models.URLField(null=True, blank=True)
def save(self, *args, **kwargs):
if self.image_url:
import urllib, os
from urlparse import urlparse
file_save_dir = self.upload_path
filename = urlparse(self.image_url).path.split('/')[-1]
urllib.urlretrieve(self.image_url, os.path.join(file_save_dir, filename))
self.image = os.path.join(file_save_dir, filename)
self.image_url = ''
super(tweet_photos, self).save()
class Pin(models.Model):
"""Pin Class"""
image_link = models.CharField(max_length=255, null=True, blank=True)
image = models.ImageField(upload_to='images/', blank=True)
title = models.CharField(max_length=255, null=True, blank=True)
source_name = models.CharField(max_length=255, null=True, blank=True)
source_link = models.CharField(max_length=255, null=True, blank=True)
description = models.TextField(null=True, blank=True)
tags = models.ForeignKey(Tag, blank=True, null=True)
def __unicode__(self):
"""Unicode class."""
return unicode(self.image_link)
def save(self, *args, **kwargs):
"""Store image locally if we have a URL"""
if self.image_link and not self.image:
result = urllib.urlretrieve(self.image_link)
self.image.save(os.path.basename(self.image_link), File(open(result[0], 'r')))
self.save()
super(Pin, self).save()
Working!
You can save image by using FileSystemStorage.
check the example below
def upload_pic(request):
if request.method == 'POST' and request.FILES['photo']:
photo = request.FILES['photo']
name = request.FILES['photo'].name
fs = FileSystemStorage()
##### you can update file saving location too by adding line below #####
fs.base_location = fs.base_location+'/company_coverphotos'
##################
filename = fs.save(name, photo)
uploaded_file_url = fs.url(filename)+'/company_coverphotos'
Profile.objects.filter(user=request.user).update(photo=photo)
class DemoImage(models.Model):
title = models.TextField(max_length=255, blank=False)
image = models.ImageField(blank=False, upload_to="images/DemoImages/")
import requests
import urllib.request
from django.core.files import File
url = "https://path/to/logo.jpg"
# Below 3 lines is to fake as browser agent
# as many sites block urllib class suspecting to be bots
opener = urllib.request.build_opener()
opener.addheaders = [("User-agent", "Mozilla/5.0")]
urllib.request.install_opener(opener)
# Issue command to actually download and create temp img file in memory
result = urllib.request.urlretrieve(url)
# DemoImage.objects.create(title="title", image=File(open(result[0], "rb")))
# ^^ This erroneously results in creating the file like
# images/DemoImages/path/to/temp/dir/logo_image_file
# as opposed to
# images/DemoImages/logo_image_file
# Solution to get the file in images/DemoImages/
reopen = open(result[0], "rb") # Returns a BufferedReader object of the temp image
django_file = File(reopen) # Create the file from the BufferedReader object
demoimg = DemoImage()
demoimg.title = "title"
demoimg.image.save("logo.png", django_file, save=True)
This approach also triggers file upload to cloudinary/S3 if so configured
So, if you have a model with an imagefield with an upload_to attribute set, such as:
class Avatar(models.Model):
image_file = models.ImageField(upload_to=user_directory_path_avatar)
then it is reasonably easy to change the image, at least in django 3.15.
In the view, when you process the image, you can obtain the image from:
self.request.FILES['avatar']
which is an instance of type InMemoryUploadedFile, as long as your html form has the enctype set and a field for avatar...
<form method="post" class="avatarform" id="avatarform" action="{% url avatar_update_view' %}" enctype="multipart/form-data">
{% csrf_token %}
<input id="avatarUpload" class="d-none" type="file" name="avatar">
</form>
Then, setting the new image in the view is as easy as the following (where profile is the profile model for the self.request.user)
profile.avatar.image_file.save(self.request.FILES['avatar'].name, self.request.FILES['avatar'])
There is no need to save the profile.avatar, the image_field already saves, and into the correct location because of the 'upload_to' callback function.
Your can use Django REST framework and python Requests library to Programmatically saving image to Django ImageField
Here is a Example:
import requests
def upload_image():
# PATH TO DJANGO REST API
url = "http://127.0.0.1:8080/api/gallery/"
# MODEL FIELDS DATA
data = {'first_name': "Rajiv", 'last_name': "Sharma"}
# UPLOAD FILES THROUGH REST API
photo = open('/path/to/photo', 'rb')
resume = open('/path/to/resume', 'rb')
files = {'photo': photo, 'resume': resume}
request = requests.post(url, data=data, files=files)
print(request.status_code, request.reason)
I save the image with uuid in django 2 python 3 because thats how django do it:
import uuid
from django.core.files import File
import urllib
httpUrl = "https://miimgeurl/image.jpg"
result = urllib.request.urlretrieve(httpUrl)
mymodel.imagefield.save(os.path.basename(str(uuid.uuid4())+".jpg"),File(open(result[0], 'rb')))
mymodel.save()
if you use admin.py you can solve the problem override (doc on django):
def save_model(self, request, obj, form, change):
obj.image_data = bytes(obj.image_name.read())
super().save_model(request, obj, form, change)
with models.py:
image_name = models.ImageField()
image_data = models.BinaryField()