Django – how to create a file and save it to a model's FileField?

Question

Here’s my model. What I want to do is generate a new file and overwrite the existing one whenever a model instance is saved:

class Kitten(models.Model):
    claw_size = ...
    license_file = models.FileField(blank=True, upload_to='license')

    def save(self, *args, **kwargs):
        #Generate a new license file overwriting any previous version
        #and update file path
        self.license_file = ???
        super(Request,self).save(*args, **kwargs)

I see lots of documentation about how to upload a file. But how do I generate a file, assign it to a model field and have Django store it in the right place?

Asked By: Greg

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Source

Answer 1

You want to have a look at FileField and FieldFile in the Django docs, and especially FieldFile.save().

Basically, a field declared as a FileField, when accessed, gives you an instance of class FieldFile, which gives you several methods to interact with the underlying file. So, what you need to do is:

self.license_file.save(new_name, new_contents)

where new_name is the filename you wish assigned and new_contents is the content of the file. Note that new_contents must be an instance of either django.core.files.File or django.core.files.base.ContentFile (see given links to manual for the details).

The two choices boil down to:

from django.core.files.base import ContentFile, File

# Using File
with open('/path/to/file') as f:
    self.license_file.save(new_name, File(f))

# Using ContentFile
self.license_file.save(new_name, ContentFile('A string with the file content'))

Answered By: tawmas

Answer 2

Accepted answer is certainly a good solution, but here is the way I went about generating a CSV and serving it from a view.

Thought it was worth while putting this here as it took me a little bit of fiddling to get all the desirable behaviour (overwrite existing file, storing to the right spot, not creating duplicate files etc).

Django 1.4.1

Python 2.7.3

#Model
class MonthEnd(models.Model):
    report = models.FileField(db_index=True, upload_to='not_used')

import csv
from os.path import join

#build and store the file
def write_csv():
    path = join(settings.MEDIA_ROOT, 'files', 'month_end', 'report.csv')
    f = open(path, "w+b")

    #wipe the existing content
    f.truncate()

    csv_writer = csv.writer(f)
    csv_writer.writerow(('col1'))

    for num in range(3):
        csv_writer.writerow((num, ))

    month_end_file = MonthEnd()
    month_end_file.report.name = path
    month_end_file.save()

from my_app.models import MonthEnd

#serve it up as a download
def get_report(request):
    month_end = MonthEnd.objects.get(file_criteria=criteria)

    response = HttpResponse(month_end.report, content_type='text/plain')
    response['Content-Disposition'] = 'attachment; filename=report.csv'

    return response

Answered By: markdsievers

Answer 3

It’s good practice to use a context manager or call close() in case of exceptions during the file saving process. Could happen if your storage backend is down, etc.

Any overwrite behavior should be configured in your storage backend. For example S3Boto3Storage has a setting AWS_S3_FILE_OVERWRITE. If you’re using FileSystemStorage you can write a custom mixin.

You might also want to call the model’s save method instead of the FileField’s save method if you want any custom side-effects to happen, like last-updated timestamps. If that’s the case, you can also set the name attribute of the file to the name of the file – which is relative to MEDIA_ROOT. It defaults to the full path of the file which can cause problems if you don’t set it – see File.__init__() and File.name.

Here’s an example where self is the model instance where my_file is the FileField / ImageFile, calling save() on the whole model instance instead of just FileField:

import os
from django.core.files import File

with open(filepath, 'rb') as fi:
    self.my_file = File(fi, name=os.path.basename(fi.name))
    self.save()

Answered By: whp

Django – how to create a file and save it to a model's FileField?

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