django - Overwriting files with default_storage.save() - django

I'm using django.core.files.storage.default_storage function for saving files, but I don't see an option where I would specify that existing ones have to be overwritten. I'm looking for some simplest and cleanest solution, so do I have to write my own function for checking if file exists, and if it does to delete it before saving?

Take a look at django-storages.
It has an overwrite storage backend that does what you're looking for. You just need to update your settings.py to use the OverwriteStorage class:
DEFAULT_FILE_STORAGE = 'storages.backends.overwrite.OverwriteStorage'

Related

Set OSM config options using API and not the global config file

When reading OSM files using GDAL, the fields that are read are defined in osmconf.ini, and if I want that certain tags don't appear within the other_tags then I need to add them to the attributes value in the corresponding sections.
This works fine, but is not really portable, so my questions is, is there a ways to define the settings saved in osmconf.ini in a portable way per project?
This is possible using the CPLSetConfigOption function, and store the config file in current working directory:
CPLSetConfigOption("OSM_CONFIG_FILE", "osmconf.ini");

If I change ImageFIeld.upload_to parameter, should it move previously uploaded files upon migration?

I have a Django model with
...
image = models.ImageField(default=None, upload_to=settings.PHOTO_UPLOAD_TO)
...
I have changed the value of settings.PHOTO_UPLOAD_TO, but the files have remained where they were, and the database entries also don't seem to have changed. A newly added image is placed in the correct new location, but not the old ones. Is it an expected behaviour? Is there a way to migrate the image locations? I would like to slightly change the directory/URL structure of my project.
The path of a FileField/ImageField is stored relative to MEDIA_ROOT, so if you want to move the files to the new upload_to folder, you're gonna need to put that in your migration yourself (like you suggested).

To use Django FileField or FilePathField?

I'm a bit confused about which field to use. What I need is just a Field that will hold a file (audio and/or another one for an image)
The FileField seems to be specifically for uploading and I think it is a bit of overkill for what I need, I also don't like how it will automatically upload and rename the files to a unique filename (file_FHjgh758.txt) everytime something is uploaded.
The problem I have with the FilePathField is that it seems to require an absolute path, which would obviously be different from dev to production, I got around this by adding this to the model...
import os
class Foo(models.Model):
path = os.path.dirname(os.path.dirname(__file__))
path = os.join(path, 'media')
audio = models.FilePathField(path=path)
I have no idea if this is safe or proper, and there aren't many examples online or in the docs to draw from.
To be clear, I just want to have a field which I can point to a file (audio or image) somewhere in my system. What would be the best way to go about doing this?
If you want to refer to files that are already on your filesystem rather than user uploaded files, then FilePathField is the one that you want.
In this regard a few comments:
Don't point to a path that is inside your app's source tree. That is not safe.
You can use settings or environment variables to handle development vs. production paths, e.g., put a FILE_PATH_FIELD_DIRECTORY setting in your development/production settings files and refer to this setting from your app:
from django.conf import settings
class Foo(models.Model):
audio = models.FilePathField(path=settings.FILE_PATH_FIELD_DIRECTORY)

Django - how to reference paths of static files and can I use them in models?

I have a django model class with several images related to each instance.
Those images follow a certain pattern and can be determined by the name field of the model.
Those Images reside within the project static files folder.
So I have written a method for my model class to generate file paths for the images. It searches the static files folder for all files that follow the pattern *.jpg (the asterisks is necessary, because the filename has incrementing numbers).
Once it has found a file it transforms the absolute filesystem path into an url that is passed to a view and template via a list.
def getImages(self)
matches[]
for filename in fnmatch.filter(
os.listdir(os.path.join(settings.STATIC_ROOT_DIR,'images')), self.name + '*.jpg'):
matches.append(
os.path.join(settings.STATIC_URL, 'images', os.path.split(filename)[1]))
return matches
This method works fine, but doesn't leave me quite satisfied. Here are the reasons:
For development mode I am required to introduce a new variable called STATIC_ROOT_DIR, to obtain the path of the static files folder. I would like to use a consistent way to reference the static root folder for development and production. How can I achieve this? I would like to avoid development mode hackery as much as possible.
I have to build a URL by joining the static_url path with other strings that will eventually make up the URL for this static file. Is there a better way to construct URLs? Maybe some library function?
Last but not least: Is it good practice to do this in a model? Or is such a task better be done by a view?
There is a STATIC_ROOT variable in settings.py. Why not use it?
Personally, I follow your way - just concatenating paths. But just found a function for that:
from django.contrib.staticfiles.templatetags.staticfiles import static
print static('yourfile.jpg')
It works for me.
I think model is a good place for it. You store files in filesystem like you store model data in database. In other words, both of these are examples of storage which is a model level thing.
I would like to throw in another one:
Using STATIC_ROOT will break if you host your files externally.
You can use the django-storage-backend yourself (untested, just written):
from django.core.files.storage import get_storage_class
from django.conf import settings
def getImages(self)
static_storage = get_storage_class(settings.STATICFILES_STORAGE)()
directories, files = static_storage.listdir('images')
return [
static_storage.url('images/' + file)
for file in files
if file.startswith(self.name) and file.endswith('.jpg')
]
This will even return the correct URL if you use CachedStaticFileStorage or S3BotoStorage (from django-storages). And this will also be fine if you are in dev-mode.

Save FileField file uploads someplace other than MEDIA_ROOT?

Can I use FileField to handle a file, but store it someplace not under MEDIA_ROOT/MEDIA_URL but somewhere else entirely.
Seeking is to ensure it is not downloadable; although denying read permissions would do the trick here's hoping for something better... like a different directory altogether.
There are a few ways to do this.
First, you can take a look at handling uploaded files from the Django docs. If you read it over, basically you can handle the upload of the file within your view in the same part where you are processing your form.
Another option, and one which I think would be better is to use a custom file storage system. You could do this very simply using the existing one as a base but simply change the location, then use it as an argument in your FileField. For example:
from django.core.files.storage import FileSystemStorage
my_store = FileSystemStorage(location='/some/other/dir')
class SomeModel(models.Model):
file = models.FileField(storage=my_store)
Hope that helps!
Considering that the only actual use of MEDIA_ROOT in all of Django is to determine where uploaded files are stored, seems like it would make more sense to just point MEDIA_ROOT where you want your uploaded files, and then use a different setting for the path to your static assets. This is the approach taken by Pinax and django-staticfiles, which use STATIC_URL and STATIC_ROOT settings.
Note that even the documentation page on serving static assets in development no longer recommends using MEDIA_ROOT for that purpose, it demonstrates using your own STATIC_DOC_ROOT setting.