I am using a file upload in my Django model like this :
def upload_path(self, filename):
return 'upload/actualities/%s/%s' % (self.id, filename)
photo = models.ImageField(upload_to=upload_path)
and my adminModel is :
from actualities.models import *
from django.contrib import admin
class ActualityAdmin(admin.ModelAdmin):
class Media:
js = ('/static/js/tiny_mce/tiny_mce.js', '/static/js/textareas.js')
admin.site.register(Actuality, ActualityAdmin)
Everything works fine except when i edit mu model because it has an id. But when I create it, the file upload happens before the model saving... So i put my file in /media/actualities/None/filename.jpg, and I want /media/2/filename.jpg
How can I force to make the file upload after the model saving?
Thank you!!!
You will probably want to override the Model's save() method, and maybe come up with a custom "don't do anything" UploadHandler, then switch back to the original one and call save again.
https://docs.djangoproject.com/en/dev/topics/http/file-uploads/
https://docs.djangoproject.com/en/dev/topics/db/models/
What I would do in this situation however, is make a custom upload handler that saves the file off into some temp space. Then I'd override the save method (in a mixin or something) that moves the file from temp to wherever you wanted it.
#Tomek's answer is also another way. If you have your model generate it's own id, then you can use that.
A second to last suggestion which is what I do with my photo blog is instead of saving all the images in a directory like media/2/filename.jpg I save the image by date uploaded. 2011/10/2/image.jpg This kind of helps any directory from getting too unwieldy.
Finally, you could hash the file names and store them in directories of hash name to kind of equally spread out the images in a directory.
I've picked the date style because that's meaningful for me with that project. Perhaps there is another way you can name an image for saving that would mean something more than "model with id 2's pics" that you could use for this problem.
Good Luck!
As workaround, try generating UUID for file name (instead of using self.id).
Related
I have class called AlbumImage
class AlbumImage(models.Model):
album = models.ForeignKey(Album, on_delete=models.PROTECT,related_name="raters")
image = models.ImageField(upload_to="Album")
when i upload first image called (for example) image1.png every thing in ok ,,,,
But if I upload another image with the same name(image1.png), and go back to the first object, I will find the second image in its place.
How do I avoid overwrite ?
The FileField and ImageField will detect this, and will add a random part to avoid overwriting.
It is only if you set it to a string, that it will take that, since then you refer to the name of a file. But if you thus work through a (Model)Form, or the ModelAdmin, or you pass it with a File object.
Indeed, if you want to save twice an item named 3.jpeg, it will generate:
/media/3.jpeg
/media/3_S12ggWF.jpeg
Im hoping someone can tell me what I am doing wrong and point me in the right direction.
So, I am creating an array of Model objects and then doing a bulk_create at the end to save them to the database. The one thing I am having an issue with is after I added the the FileField I cannot exactly how I need to associate data with that field. The files don't end up in the upload_to folder set nor do they end up being associated with the record itself. Id also add that I am using PyPDf2 to create the PDF files before trying to associate them to an instance of my model.
So to give you an idea on what I am trying to do. I am running this code to create the PDFs initially.
if pdf_file_name is not None:
num_pages_current_doc = page['pageRange'][-1]
input_pdf = PdfFileReader(open(pdf_file_name, 'rb'))
output_pdf = PdfFileWriter()
for _ in range(num_pages_current_doc):
output_pdf.addPage(input_pdf.getPage(page_counter))
page_counter += 1
with open(str(uuid.uuid4())+'.pdf', 'wb') as output:
output_pdf.write(output)
logging.info(f'Wrote filename: { output.name }')
The saved file I then want to associated with a model instance below this, the code looks something like this:
document = document(
location=location,
Field2=Field2, etc etc .....
pdf = ???
Im unsure how to set the field for that pdf part, Ive tried using the File() method on it. Tried putting just output.name for the field, Im not sure how to go about making this work.
Could anyone give me some insight?
Thanks!
See the FieldFile.save(name, content, save=True) method in django docs. https://docs.djangoproject.com/en/3.2/ref/models/fields/#filefield-and-fieldfile
I have a Django app where users can upload images and can have a processed version of the images if they want. and the processing function returns the path, so my approach was
model2.processed_image = processingfunction( model1.uploaded_image.path)
and as the processing function returns path here's how it looks in my admin view
not like the normally uploaded images
In my machine it worked correctly and I always get a 404 error for the processed ones while the normally uploaded is shown correctly when I try to change the url of the processed from
myurl.com/media/home/ubuntu/Eyelizer/media/path/to/the/image
to
myurl.com/media/path/to/the/image
so how can I fix this ? is there a better approach to saving the images manually to the database ?
I have the same function but returns a Pil.image.image object and I've tried many methods to save it in a model but I didn't know how so I've made the function return a file path.
I think the problem is from nginx where I define the media path.
should/can I override the url attribute of the processedimage?
making something like
model.processed_image.url = media/somefolder/filename
Instead of using the PIL Image directly, create a django.core.files.File.
Example:
from io import BytesIO
from django.core.files import File
img_io = BytesIO() # create a BytesIO object to temporarily save the file in memory
img = processingfunction( model1.uploaded_image.path)
img.save(img_io, 'PNG') # save the PIL image to the BytesIO object
img_file = File(thumb_io, name='some-name.png') # create the File object
# you can use the `name` from `model1.uploaded_image` and use
# that above
# finally, pass the image file to your model field
model2.processed_image = img_file
To avoid repetition of this code, it would be a good idea to keep this code in processingfunction and return the File object directly from there.
My approach is a bit different from #Xyres's, I thought xyres's would make a duplicate of the existing image and create a new one and when I tried overriding the URL attribute it returned an error of
can't set the attribute
but when I saw this question and this ticket I tried making this and it worked
model2.processed_image = processingfunction(model1.uploaded_image.path)
full_path = model2.processed_image.path
model2.processed_image.name = full_path.split('media')[1]
so that explicitly making the URL media/path/to/image and cut out all of the unneeded parts like home/ubuntu and stuff
I would like to keep the original file name of an UploadedFile in Django that has its location stored in a FileField. Right now I am observing that if two files have the same name, the first file uploaded keeps its original name but the second time a file with that name is uploaded, it has a random string appended to make the file name unique. One solution is to add an additional field to the model: Django: How to save original filename in FileField? or Saving Original File Name in Django with FileField but these solutions seem suboptimal as they require changing the Model fields.
An alternative would be to prepend a random directory path to the front of the file make sure that in a given directory the file name is unique and allowing the basename to remain unchanged. One way to do this would be to pass in a callable upload_to that does just that. Another option would be to subclass FileField and override get_filename to not strip the input filename to the basename allowing the caller to pass in a filename with a prepended path. The latter option is not ideal if I want to use an ImageField as I would have to subclass that as well.
In looking at the code that actually generates the unique filename by appending the random string, it looks like the best solution to this problem might be to subclass the Storage class in-use and override get_available_name method to create unique filenames by prepending a directory rather than post-pending the string to the base name.
Sorry for the quick answere, here is another approach to your question :
The idea here is to create an unique folder for each uploaded file.
# in your settings.py file
MY_FILE_PATH = 'stored_files/'
The path were your files will be stored : /public/media/stored_files
# somewhere in your project create an utils.py file
import random
try:
from hashlib import sha1 as sha_constructor
except ImportError:
from django.utils.hashcompat import sha_constructor
def generate_sha1(string, salt=None):
"""
Generates a sha1 hash for supplied string.
:param string:
The string that needs to be encrypted.
:param salt:
Optionally define your own salt. If none is supplied, will use a random
string of 5 characters.
:return: Tuple containing the salt and hash.
"""
if not isinstance(string, (str, unicode)):
string = str(string)
if isinstance(string, unicode):
string = string.encode("utf-8")
if not salt:
salt = sha_constructor(str(random.random())).hexdigest()[:5]
hash = sha_constructor(salt+string).hexdigest()
return (salt, hash)
In your models.py
from django.conf import settings
from utils.py import generate_sha1
def upload_to_unqiue_folder(instance, filename):
"""
Uploads a file to an unique generated Path to keep the original filename
"""
salt, hash = generate_sha1('{}{}'.format(filename, get_datetime_now().now))
return '%(path)s%(hash_path)s%(filename)s' % {'path': settings.MY_FILE_PATH,
'hash_path': hash[:10],
'filename': filename}
#And then add in your model fileField the uplaod_to function
class MyModel(models.Model):
file = models.FileField(upload_to=upload_to_unique_folder)
The file will be uploaded to this location :
public/media/stored_file_path/unique_hash_folder/my_file.extention
Note : I got the code from Django userena sources, and adapted it to my needs
Note2 : For more informations take a look at this greate post on Django File upload : File upload example
Have a good day.
Edit : Trying to provide a working solution :)
To my understanding, during the form submission/file upload process, you can add form validation functions.
During the validation and cleaning process, you could check that the database does not already have a duplicate name (ie. query to see if that file name exists).
If it is duplicate, you could just rename it xyz_1, xyz_2, etc
I have an API endpoint with Django Rest Framework to upload an image.
class MyImageSerializer(serializers.ModelSerializer):
image = serializers.ImageField(source='image')
I can upload images but they are saved with the filename that is sent from the client which can result to collisions. I would like instead to upload the file to my CDN with a timestamp filename.
Generating the filename is not the problem, just saving the image with it.
Any one knows how to do that?
Thanks.
If your image is of type ImageField from django, then you don't really have to do anything, not even declare it in your serializer like you did. It's enough to add it in the fields attribute and django will handle collisions. This means django will add _index on each new file which might generate a collision, so if you upload a file named 'my_pic.jpg' 5 times, you will actually have files 'my_pic.jpg', 'my_pic_1.jpg', 'my_pic_2.jpg', 'my_pic_3.jpg', 'my_pic_4.jpg' on your server.
Now, this is done using django's implementation for FileSystemStorage (see here), but if you want it to append a timestamp to your filename, all you have to do is write a storage class where you overwrite the get_available_name(name) method. Example:
class MyFileSystemStorage(FileSystemStorage):
def get_available_name(self, name):
''' name is the current file name '''
now = time.time()
stamp = datetime.datetime.fromtimestamp(now).strftime('%Y-%m-%d-%H-%M-%S')
return '{0}_{1}'.format(name, str(stamp))
And the image field in your model:
image = models.ImageField(upload_to='your upload dir', storage= MyFileSystemStorage)
Important update
As of August 20, 2014 this is no longer an issue, since Django found a vulnerability related to this behaviour (thanks #mlissner for pointing it out) . Important excerpt :
We’ve remedied the issue by changing the algorithm for generating file
names if a file with the uploaded name already exists.
Storage.get_available_name() now appends an underscore plus a random 7
character alphanumeric string (e.g. "_x3a1gho"), rather than iterating
through an underscore followed by a number (e.g. "_1", "_2", etc.).