Develop Reference

Python: Getting local file path from request.files

I need help uploading a file directly from an HTML form to an API. I've seen this being done for remote URLs, but I don't know how to do this for local files? I tried writing this, but its not working:
uploadmedia = request.files['fileupload']
client = Client('thisismykey')
with open(uploadmedia, 'rb') as file:
new_upload = client.uploads('<space-id>').create(file)
The line client.uploads is what is specified in the API docs here. I just need to be able to get the file path.
The comments suggest the following:
# you can use either a file-like object or a path.
# If you use a path, the SDK will open it, create the upload and
# close the file afterwards.
I am assuming that request.files['fileupload'] is a file like object, so I just passed that along.
The above code gives me the following error:
File "D:\Gatsby\submission\flask-tailwindcss-starter\app\__init__.py", line 28, in index
with open(uploadmedia, 'rb') as file:
TypeError: expected str, bytes or os.PathLike object, not FileStorage
I know that in this example, uploadmedia.filename would get me the file's name, but what is the attribute for the path? How do I do that?

The request.files['file'] is an instance of a FileStorage class. refer to api, you cannot use with open(uploadmedia, 'rb') as file: .
try using stream attribute :
uploadmedia = request.files['fileupload']
client = Client('thisismykey')
new_upload = client.uploads('<space-id>').create(uploadmedia.stream)

How to import pickled python object into Django view.py file

This file runs perfectly well on its own:
import pickle
with open('model.obj', 'rb') as f:
model = pickle.load(f)
print(model)
But what I really want to do is to use model, which is a very large stored python dictionary, in my Django views.py file, like so:
from django.shortcuts import render
from modeler.load_model import model
def page(request):
return render(request, 'page.html', {'model':model})
However, when I try this, I get the error below:
File "C:\Users\mmm\PycharmProjects\MyProject\modeler\load_model.py", line 3, in <module>
with open('model.obj', 'rb') as f:
FileNotFoundError: [Errno 2] No such file or directory: 'model.obj'
I don't understand why, because if I run the first file directly it works fine - the only extra step here is importing it into views.py (I tried including the entire load_model.py code within views.py at first, but got this error, so then tried it in this separate file and it worked on its own, so obviously the issue is that I don't know how to correctly load/import my python model objects or files within Django.

I found the answer! Apparently, all I had to do was to put the absolute path to 'model.obj' in my open statement, not just the relative path, like this:
with open('C:\\Users\\mmm\\PycharmProjects\\MyProject\\modeler\\model.obj', 'rb') as f:
model = pickle.load(f)

Django - create CSV on the fly and save it to filesystem and model file field

In my django app, I have a management command which I want to use to process some data and create a model instance as well as file in filesystem and save its path to above instance's file field.
My management command:
import datetime
import timy
from django.core.management.base import BaseCommand
import tempfile
from core.models import Entry
from np_web.models import Dump
import calendar
import csv
class Command(BaseCommand):
help = 'Create dump and file for current day'
def handle(self, *args, **options):
with timy.Timer() as timer:
today = datetime.date.today()
dump_title = '{}.{}.{}.csv'.format(today.day, today.month, today.year)
entries = Entry.objects.all()
dump = Dump.objects.create(all_entries=entries.count())
dump.save()
print('Dump created with uuid:', dump.uuid)
print('Now create data file for dump')
with tempfile.NamedTemporaryFile() as temp_csv:
writer = csv.writer(temp_csv)
writer.writerow([
'Column1',
'Column2',
])
for entry in entries:
writer.writerow([
entry.first_name,
entry.last_name
])
dump.file.save(dump_title, temp_csv)
My Dump model:
class Dump(BaseModel):
created = models.DateField(auto_now_add=True)
all_entries = models.PositiveIntegerField()
file = models.FileField(verbose_name=_('Attachment'), upload_to=dump_file_upload_path, max_length=2048,
storage=attachment_upload_storage, null=True, blank=True)
Anyway, it doesn't work. It is throwing an error:
TypeError: a bytes-like object is required, not 'str'
I am also not sure if using temporary file is a best solution out there.

A few comments to hopefully direct you to a solution.
For starters, "If delete is true (the default), the file is deleted as soon as it is closed.", and assuming that the spacing you have is correct, you are closing the file prior to attempting to save the file. This would result in the file being empty.
My suggestion would be to simply create the file like normal, and delete it afterwards. A simple approach (though there may be a better) would be to create the file, save it, and then remove the original (temp) copy.
In addition, when saving the file, you want to save it using Django's File wrapper.
So you could do something like the following:
from django.core.files import File
import os
with open(temp, 'rb') as f:
doc_file = File(f)
dump.file.save("filename", doc_file, True)
dump_file.save()
try:
os.remove(temp)
except Expection:
print('Unable to remove temp file')

changing file name in django

I have a this model...
class MyModel(models.Model):
...
file = models.FileField(upload_to='files/',null=True, blank=True)
...
when i upload a file, example file name is docfile.doc. when i change the file or i rewrite it and upload again docfile.doc the file will become docfile_1.doc and the old docfile.doc is still exist.
i am doing the uploading and saving data in django-admin
my question is, how can i remove the old docfile.doc if i upload the new docfile.doc and the file name is still docfile.doc?
can anyone help me in my case? thanks in advance
i try this one :
def content_file_name(instance, filename):
print instance
print filename
file = os.path.exists(filename)
print file
if file:
os.remove(filename)
return "file/"+str(filename)
class MyModel(models.Model):
...
file = models.FileField(upload_to=content_file_name,null=True, blank=True)
...
but nothing happend, when i upload docfile.doc again, it will become docfile_1.doc and the old docfile.doc still exist.

i got it... i use this
def content_file_name(instance, filename):
print instance
print filename
file = os.path.exists("media/file/"+str(filename))
print file
if file:
os.remove("media/file/"+str(filename))
return "file/"+str(filename)

I don't know exactly how to do it, but i think these links can help you:
Here you can find the two options that a FileField accept. The one that i think will interest you the most is FileField.storage. You can pass a storage object in that parameter.
It says:
FileField.storage: Optional. A storage object, which handles the storage and retrieval of your files.
Then, if you read this you would see that you can write your own storage object. Here is some explanation on how to do it. I think that you could just override the _save method in order to accomplish what you want to do (i.e: if the file already exists, remove it before saving the new copy.)
But be careful! I don't know which is the source of the files you are going to store. Maybe, your app is going to recieve lots of files with the same name, although they are all different. In this case, you would want to use a callable as the FileField.upload_to parameter, so that determine a unique filename for each file your site recieve.
I hope this helps you!

You could also have a look here: ImageField overwrite image file with same name
Define your own storage and overwrite its get available_name method.

The next code solves your problem. You override pre_save method where image is actually saved to storage. Please, rename functions for your project. Use newly created image field ImageFieldWithPermantName with your upload_to function (content_file_name).
If the code is too complicated you could simplify it. I use the code to do more complex operations for uploading images: I create thumbnails on-the-fly in custom _save_image function. So, you can simplify it.
from PIL import Image
import StringIO
from django.db.models import ImageField
from django.db.models.fields.files import FileField
from dargent.settings import MEDIA_ROOT
import os
class ImageFieldWithPermanentName( ImageField ):
def pre_save( self, model_instance, add ):
file = super( FileField, self ).pre_save(model_instance, add)
if file and not file._committed:
if callable( self.upload_to ):
path = self.upload_to( model_instance, "" )
else:
path = self.upload_to
file.name = path # here we set the same name to a file
path = os.path.join( MEDIA_ROOT, path )
chunks = _get_chunks( file.chunks() )
_save_image( chunks, path )
return file
def _get_chunks( chunks ):
chunks_ = ""
for chunk in chunks:
chunks_ += chunk
return chunks_
def _get_image( chunks ):
chunks_ = ""
for chunk in chunks:
chunks_ += chunk
virt_file = StringIO.StringIO( chunks_ )
image = Image.open( virt_file )
return image
def _save_image( chunks, out_file_path ):
image = _get_image( chunks )
image.save( out_file_path, "JPEG", quality = 100 )

ZipExtFile to Django File

I am wondering whether there is a way to upload a zip file to django web server and put the zip's files into django database WITHOUT accessing the actual file system in the process (e.g. extracting the files in the zip into a tmp dir and then load them)
Django provides a function to convert python File to Django File, so if there is a way to convert ZipExtFile to python File, it should be fine.
thanks for help!
Django model:
from django.db import models
class Foo:
file = models.FileField(upload_to='somewhere')
Usage:
from zipfile import ZipFile
from django.core.exceptions import ValidationError
from django.core.files import File
from io import BytesIO
z = ZipFile('zipFile')
istream = z.open('subfile')
ostream = BytesIO(istream.read())
tmp = Foo(file=File(ostream))
try:
tmp.full_clean()
except Validation, e:
print e
Output:
{'file': [u'This field cannot be blank.']}
[SOLUTION] Solution using an ugly hack:
As correctly pointed out by Don Quest, file-like classes such as StringIO or BytesIO should represent the data as a virtual file. However, Django File's constructor only accepts the build-in file type and nothing else, although the file-like classes would have done the job as well. The hack is to set the variables in Django::File manually:
buf = bytesarray(OPENED_ZIP_OBJECT.read(FILE_NAME))
tmp_file = BytesIO(buf)
dummy_file = File(tmp_file) # this line actually fails
dummy_file.name = SOME_RANDOM_NAME
dummy_file.size = len(buf)
dummy_file.file = tmp_file
# dummy file is now valid
Please keep commenting if you have a better solution (except for custom storage)

There's an easier way to do this:
from django.core.files.base import ContentFile
uploaded_zip = zipfile.ZipFile(uploaded_file, 'r') # ZipFile
for filename in uploaded_zip.namelist():
with uploaded_zip.open(filename) as f: # ZipExtFile
my_django_file = ContentFile(f.read())
Using this, you can convert a file that was uploaded to memory directly to a django file. For a more complete example, let's say you wanted to upload a series of image files inside of a zip to the file system:
# some_app/models.py
class Photo(models.Model):
image = models.ImageField(upload_to='some/upload/path')
...
# Upload code
from some_app.models import Photo
for filename in uploaded_zip.namelist():
with uploaded_zip.open(filename) as f: # ZipExtFile
new_photo = Photo()
new_photo.image.save(filename, ContentFile(f.read(), save=True)

Without knowing to much about Django, i can tell you to take a look at the "io" package.
You could do something like:
from zipfile import ZipFile
from io import StringIO
zname,zipextfile = 'zipcontainer.zip', 'file_in_archive'
istream = ZipFile(zname).open(zipextfile)
ostream = StringIO(istream.read())
And then do whatever you would like to do with your "virtual" ostream Stream/File.

I've used the following django file class to avoid the need to read ZipExtFile into a another datastructure (StingIO or BytesIO) while properly impelementing what Django needs in order to save the file directly.
from django.core.files.base import File
class DjangoZipExtFile(File):
def __init__(self, zipextfile, zipinfo):
self.file = zipextfile
self.zipinfo = zipinfo
self.mode = 'r'
self.name = zipinfo.filename
self._size = zipinfo.file_size
def seek(self, position):
if position != 0:
#this will raise an unsupported operation
return self.file.seek(position)
#TODO if we have already done a read, reopen file
zipextfile = archive.open(path, 'r')
zipinfo = archive.getinfo(path)
djangofile = DjangoZipExtFile(zipextfile, zipinfo)
storage = DefaultStorage()
result = storage.save(djangofile.name, djangofile)