I am trying to catch a file sent with form and perform some operations on it before it will be saved. So I need to create a copy of this file in temp directory, but I don't know how to reach it. Shutil's functions fail to copy this file, since there is no path to it. So is there a way to do this operation in some other way ?
My code :
image = form.cleaned_data['image'] temp = os.path.join(settings.PROJECT_PATH, 'tmp') sourceFile = image.name # without .name here it wasn't working either import shutil shutil.copy(sourceFile, temp)
Which raises :
Exception Type: IOError at /
Exception Value: (2, 'No such file or directory')
And the debug :
# (..)\views.py in function 67. sourceFile = image.name 68. import shutil 69. shutil.copy2(sourceFile, temp) ... # (..)\Python26\lib\shutil.py in copy2 92. """Copy data and all stat info ("cp -p src dst"). 93. 94. The destination may be a directory. 95. 96. """ 97. if os.path.isdir(dst): 98. dst = os.path.join(dst, os.path.basename(src)) 99. copyfile(src, dst) ... 100. copystat(src, dst) 101. ▼ Local vars Variable Value dst u'(..)\\tmp\\myfile.JPG' src u'myfile.JPG' # (..)\Python26\lib\shutil.py in copyfile 45. """Copy data from src to dst""" 46. if _samefile(src, dst): 47. raise Error, "`%s` and `%s` are the same file" % (src, dst) 48. 49. fsrc = None 50. fdst = None 51. try: 52. fsrc = open(src, 'rb') ... 53. fdst = open(dst, 'wb') 54. copyfileobj(fsrc, fdst) 55. finally: 56. if fdst: 57. fdst.close() 58. if fsrc: ▼ Local vars Variable Value dst u'(..)\\tmp\\myfile.JPG' fdst None fsrc None src u'myfile.JPG'
This is similar question, it might help.
import os from django.core.files.storage import default_storage from django.core.files.base import ContentFile from django.conf import settings data = request.FILES['image'] # or self.files['image'] in your form path = default_storage.save('tmp/somename.mp3', ContentFile(data.read())) tmp_file = os.path.join(settings.MEDIA_ROOT, path)
As mentioned by @ups, when uploading large files, you don't want to clog up system memory with a
From Django docs :
UploadedFile.chunks()instead of using
read()ensures that large files don't overwhelm your system's memory
from django.core.files.storage import default_storage filename = "whatever.xyz" # received file name file_obj = request.data['file'] with default_storage.open('tmp/'+filename, 'wb+') as destination: for chunk in file_obj.chunks(): destination.write(chunk)
This will save the file at
MEDIA_ROOT/tmp/ as your
default_storage will unless told otherwise.