I'm trying to generate dynamic file paths in django. I want to make a file system like this:
-- user_12
--- photo_1
--- photo_2
--- user_ 13
---- photo_1
I found a related question : Django Custom image upload field with dynamic path
Here, they say we can change the upload_to path and leads to https://docs.djangoproject.com/en/stable/topics/files/ doc. In the documentation, there is an example :
from django.db import models
from django.core.files.storage import FileSystemStorage
fs = FileSystemStorage(location='/media/photos')
class Car(models.Model):
...
photo = models.ImageField(storage=fs)
But, still this is not dynamic, I want to give Car id to the image name, and I cant assign the id before Car definition completed. So how can I create a path with car ID ??
You can use a callable in the upload_to argument rather than using custom storage. See the docs, and note the warning there that the primary key may not yet be set when the function is called. This can happen because the upload may be handled before the object is saved to the database, so using ID might not be possible. You might want to consider using another field on the model such as slug. E.g:
import os
def get_upload_path(instance, filename):
return os.path.join(
"user_%d" % instance.owner.id, "car_%s" % instance.slug, filename)
then:
photo = models.ImageField(upload_to=get_upload_path)
You can use lambda function as below, take note that if instance is new then it won't have the instance id, so use something else:
logo = models.ImageField(upload_to=lambda instance, filename: 'directory/images/{0}/{1}'.format(instance.owner.id, filename))
https://docs.djangoproject.com/en/stable/ref/models/fields/#django.db.models.FileField.upload_to
def upload_path_handler(instance, filename):
return "user_{id}/{file}".format(id=instance.user.id, file=filename)
class Car(models.Model):
...
photo = models.ImageField(upload_to=upload_path_handler, storage=fs)
There is a warning in the docs, but it shouldn't affect you since we're after the User ID and not the Car ID.
In most cases, this object will not
have been saved to the database yet,
so if it uses the default AutoField,
it might not yet have a value for its
primary key field.
My solution is not elegant, but it works:
In the model, use a the standard function that will need the id/pk
def directory_path(instance, filename):
return 'files/instance_id_{0}/{1}'.format(instance.pk, filename)
in views.py save the form like this:
f=form.save(commit=False)
ftemp1=f.filefield
f.filefield=None
f.save()
#And now that we have crated the record we can add it
f.filefield=ftemp1
f.save()
It worked for me.
Note: My filefield in models and allowed for Null values. Null=True
Well very late to the party but this one works for me.
def content_file_name(instance, filename):
upload_dir = os.path.join('uploads',instance.albumname)
if not os.path.exists(upload_dir):
os.makedirs(upload_dir)
return os.path.join(upload_dir, filename)
Model like this only
class Album(models.Model):
albumname = models.CharField(max_length=100)
audiofile = models.FileField(upload_to=content_file_name)
There are two solutions on DjangoSnippets
Two-stage save: https://djangosnippets.org/snippets/1129/
Prefetch the ID (PostgreSQL only): https://djangosnippets.org/snippets/2731/
This guy has a way to do dynamic path. The idea is to set your favourite storage and customise "upload_to()" parameter with a function.
Hope this helps.
I found out a different solution, which is dirty, but it works. You should create a new dummy model, which is self synchronized with the original one. I'm not proud of this, but didn't find another solution. In my case I want to upload files, and store each in a directory named after the model id (because I'll generate there more files).
the model.py
class dummyexperiment(models.Model):
def __unicode__(self):
return str(self.id)
class experiment(models.Model):
def get_exfile_path(instance, filename):
if instance.id == None:
iid = instance.dummye.id
else:
iid = instance.id
return os.path.join('experiments', str(iid), filename)
exfile = models.FileField(upload_to=get_exfile_path)
def save(self, *args, **kwargs):
if self.id == None:
self.dummye = dummyexperiment()
self.dummye.save()
super(experiment, self).save(*args, **kwargs)
I'm very new in python and in django, but it seems like ok for me.
another solution:
def get_theme_path(instance, filename):
id = instance.id
if id == None:
id = max(map(lambda a:a.id,Theme.objects.all())) + 1
return os.path.join('experiments', str(id), filename)
As the primary key (id) may not be available if the model instance was not saved to the database yet, I wrote my FileField subclasses which move the file on model save, and a storage subclass which removes the old files.
Storage:
class OverwriteFileSystemStorage(FileSystemStorage):
def _save(self, name, content):
self.delete(name)
return super()._save(name, content)
def get_available_name(self, name):
return name
def delete(self, name):
super().delete(name)
last_dir = os.path.dirname(self.path(name))
while True:
try:
os.rmdir(last_dir)
except OSError as e:
if e.errno in {errno.ENOTEMPTY, errno.ENOENT}:
break
raise e
last_dir = os.path.dirname(last_dir)
FileField:
def tweak_field_save(cls, field):
field_defined_in_this_class = field.name in cls.__dict__ and field.name not in cls.__bases__[0].__dict__
if field_defined_in_this_class:
orig_save = cls.save
if orig_save and callable(orig_save):
assert isinstance(field.storage, OverwriteFileSystemStorage), "Using other storage than '{0}' may cause unexpected behavior.".format(OverwriteFileSystemStorage.__name__)
def save(self, *args, **kwargs):
if self.pk is None:
orig_save(self, *args, **kwargs)
field_file = getattr(self, field.name)
if field_file:
old_path = field_file.path
new_filename = field.generate_filename(self, os.path.basename(old_path))
new_path = field.storage.path(new_filename)
os.makedirs(os.path.dirname(new_path), exist_ok=True)
os.rename(old_path, new_path)
setattr(self, field.name, new_filename)
# for next save
if len(args) > 0:
args = tuple(v if k >= 2 else False for k, v in enumerate(args))
kwargs['force_insert'] = False
kwargs['force_update'] = False
orig_save(self, *args, **kwargs)
cls.save = save
def tweak_field_class(orig_cls):
orig_init = orig_cls.__init__
def __init__(self, *args, **kwargs):
if 'storage' not in kwargs:
kwargs['storage'] = OverwriteFileSystemStorage()
if orig_init and callable(orig_init):
orig_init(self, *args, **kwargs)
orig_cls.__init__ = __init__
orig_contribute_to_class = orig_cls.contribute_to_class
def contribute_to_class(self, cls, name):
if orig_contribute_to_class and callable(orig_contribute_to_class):
orig_contribute_to_class(self, cls, name)
tweak_field_save(cls, self)
orig_cls.contribute_to_class = contribute_to_class
return orig_cls
def tweak_file_class(orig_cls):
"""
Overriding FieldFile.save method to remove the old associated file.
I'm doing the same thing in OverwriteFileSystemStorage, but it works just when the names match.
I probably want to preserve both methods if anyone calls Storage.save.
"""
orig_save = orig_cls.save
def new_save(self, name, content, save=True):
self.delete(save=False)
if orig_save and callable(orig_save):
orig_save(self, name, content, save=save)
new_save.__name__ = 'save'
orig_cls.save = new_save
return orig_cls
#tweak_file_class
class OverwriteFieldFile(models.FileField.attr_class):
pass
#tweak_file_class
class OverwriteImageFieldFile(models.ImageField.attr_class):
pass
#tweak_field_class
class RenamedFileField(models.FileField):
attr_class = OverwriteFieldFile
#tweak_field_class
class RenamedImageField(models.ImageField):
attr_class = OverwriteImageFieldFile
and my upload_to callables look like this:
def user_image_path(instance, filename):
name, ext = 'image', os.path.splitext(filename)[1]
if instance.pk is not None:
return os.path.join('users', os.path.join(str(instance.pk), name + ext))
return os.path.join('users', '{0}_{1}{2}'.format(uuid1(), name, ext))
MEDIA_ROOT/
/company_Company1/company.png
/shop_Shop1/shop.png
/bikes/bike.png
def photo_path_company(instance, filename):
# file will be uploaded to MEDIA_ROOT/company_<name>/
return 'company_{0}/{1}'.format(instance.name, filename)
class Company(models.Model):
name = models.CharField()
photo = models.ImageField(max_length=255, upload_to=photo_path_company)
def photo_path_shop(instance, filename):
# file will be uploaded to MEDIA_ROOT/company_<name>/shop_<name>/
parent_path = instance.company._meta.get_field('photo').upload_to(instance.company, '')
return parent_path + 'shop_{0}/{1}'.format(instance.name, filename)
class Shop(models.Model):
name = models.CharField()
photo = models.ImageField(max_length=255, upload_to=photo_path_shop)
def photo_path_bike(instance, filename):
# file will be uploaded to MEDIA_ROOT/company_<name>/shop_<name>/bikes/
parent_path = instance.shop._meta.get_field('photo').upload_to(instance.shop, '')
return parent_path + 'bikes/{0}'.format(filename)
class Bike(models.Model):
name = models.CharField()
photo = models.ImageField(max_length=255, upload_to=photo_path_bike)
You can override model's save method:
def save_image(instance, filename):
instance_id = f'{instance.id:03d}' # 001
return f'{instance_id}-{filename.lower()}' # 001-foo.jpg
class Resource(models.Model):
photo = models.ImageField(upload_to=save_image)
def save(self, *args, **kwargs):
if self.id is None:
photo = self.photo
self.photo = None
super().save(*args, **kwargs)
self.photo = photo
if 'force_insert' in kwargs:
kwargs.pop('force_insert')
super().save(*args, **kwargs)
The method will be
def user_directory_path(field_name):
def upload_path(instance, filename):
year = datetime.now().year
name, ext = instance.user, os.path.splitext(filename)[1]
return f'photos/{year}/{instance._meta.model_name}s/{instance.user}/{field_name}_{name}{ext}'
return upload_path
And in your models you can have as many ImageField as you like. example
photo = models.ImageField(upload_to=user_directory_path('photo'), null=True, blank=True,)
passport_photo = models.ImageField(upload_to=user_directory_path('passport_photo'), null=True, blank=True,)
Related
I'm trying to set up my uploads so that if user joe uploads a file it goes to MEDIA_ROOT/joe as opposed to having everyone's files go to MEDIA_ROOT. The problem is I don't know how to define this in the model. Here is how it currently looks:
class Content(models.Model):
name = models.CharField(max_length=200)
user = models.ForeignKey(User)
file = models.FileField(upload_to='.')
So what I want is instead of '.' as the upload_to, have it be the user's name.
I understand that as of Django 1.0 you can define your own function to handle the upload_to but that function has no idea of who the user will be either so I'm a bit lost.
Thanks for the help!
You've probably read the documentation, so here's an easy example to make it make sense:
def content_file_name(instance, filename):
return '/'.join(['content', instance.user.username, filename])
class Content(models.Model):
name = models.CharField(max_length=200)
user = models.ForeignKey(User)
file = models.FileField(upload_to=content_file_name)
As you can see, you don't even need to use the filename given - you could override that in your upload_to callable too if you liked.
This really helped. For a bit more brevity's sake, decided to use lambda in my case:
file = models.FileField(
upload_to=lambda instance, filename: '/'.join(['mymodel', str(instance.pk), filename]),
)
A note on using the 'instance' object's pk value. According to the documentation:
In most cases, this object will not have been saved to the database yet, so if it uses the default AutoField, it might not yet have a value for its primary key field.
Therefore the validity of using pk depends on how your particular model is defined.
If you have problems with migrations you probably should be using #deconstructible decorator.
import datetime
import os
import unicodedata
from django.core.files.storage import default_storage
from django.utils.deconstruct import deconstructible
from django.utils.encoding import force_text, force_str
#deconstructible
class UploadToPath(object):
def __init__(self, upload_to):
self.upload_to = upload_to
def __call__(self, instance, filename):
return self.generate_filename(filename)
def get_directory_name(self):
return os.path.normpath(force_text(datetime.datetime.now().strftime(force_str(self.upload_to))))
def get_filename(self, filename):
filename = default_storage.get_valid_name(os.path.basename(filename))
filename = force_text(filename)
filename = unicodedata.normalize('NFKD', filename).encode('ascii', 'ignore').decode('ascii')
return os.path.normpath(filename)
def generate_filename(self, filename):
return os.path.join(self.get_directory_name(), self.get_filename(filename))
Usage:
class MyModel(models.Model):
file = models.FileField(upload_to=UploadToPath('files/%Y/%m/%d'), max_length=255)
I wanted to change the upload path in runtime, and none of the solutions were suitable for this need.
this is what I've done:
class Content(models.Model):
name = models.CharField(max_length=200)
user = models.ForeignKey(User)
file = models.FileField(upload_to=DynamicUploadPath.get_file_path)
class ContentSerializer(serializers.ModelSerializer):
class Meta:
model = Content
fields = '__all__'
class UploadDir(models.TextChoices):
PRODUCT = 'PRD', _('Product')
USER_PROFILE = 'UP', _('User Profile')
class DynamicUploadPath:
dir: UploadDir = None
#classmethod
def get_file_path(cls, instance, filename):
return str(cls.dir.name.lower() + '/' + filename)
def set_DynamicUploadPath(dir: UploadDir):
DynamicUploadPath.dir = dir
class UploadFile(APIView):
parser_classes = (MultiPartParser, FormParser)
def post(self, request):
# file save path: MEDIA_ROOT/product/filename
set_DynamicUploadPath(UploadDir.PRODUCT)
# file save path: MEDIA_ROOT/user_profile/filename
# set_DynamicUploadPath(UploadDir.USER_PROFILE)
serializer = ContentSerializer(data=request.data)
serializer.is_valid(raise_exception=True)
serializer.save()
return Response(serializer.data, status=status.HTTP_200_OK)
If you have a user instance, let there be a quick setup to generate
<model-slug>/<username>-<first_name>-<last_name>/filename-random.png
eg:
/medias/content/ft0004-john-doe/filename-lkl9237.png
def upload_directory_name(instance, filename):
user = getattr(instance, 'user', None)
if user:
name = f"{user.username}-{user.get_full_name().replace(' ', '-')}"
else:
name=str(instance)
model_name = instance._meta.verbose_name.replace(' ', '-')
return str(os.path.pathsep).join([model_name, name, filename])
class Content(models.Model):
name = models.CharField(max_length=200)
user = models.ForeignKey(User)
file = models.FileField(upload_to=upload_directory_name)
[A Modified Version of #SmileyChris ]
I have a form which gets current logged in user, some inputs and a file:
class AddItemForm(ModelForm):
class Meta:
model = Item
exclude = ['user']
For this form a have a view:
item_form = AddItemForm(request.POST, request.FILES)
if item_form.is_valid():
item = item_form.save(commit=False)
item.user = request.user
item.save()
for this item's file field i am using upload_to feature. here is my modal:
class Item(models.Model):
user = models.ForeignKey(User)
cover_image = models.FileField(upload_to=get_upload_path)
def get_upload_path(instance, filename):
return "items/user_{user_id}/item_{item_id}/{filename}".format(user_id=instance.user.id, item_id=instance.id,filename=filename)
problem is that i can't see the current instance id in uploaded path because of following line:
item = item_form.save(commit=False)
it has not instance id yet and instead of current item id it create user_1/item_NONE/file
how can i set id to this path?
thanks in advance
Here I've found idea && code based on using the post_save signal, when object created move from temp directory to specified one in model class:
use_key and upload_to are optional. use_key defaults to False. If it is True then the id of the instance will be used as a prefix for the new file as there is the potential for overwriting now that we are moving the file. upload_to will simply define the temporary directory to upload the files to initially.
from django.db.models import ImageField, FileField, signals
from django.dispatch import dispatcher
from django.conf import settings
import shutil, os, glob, re
from distutils.dir_util import mkpath
class CustomImageField(ImageField):
"""Allows model instance to specify upload_to dynamically.
Model class should have a method like:
def get_upload_to(self, attname):
return 'path/to/{0}'.format(self.id)
"""
def __init__(self, *args, **kwargs):
kwargs['upload_to'] = kwargs.get('upload_to', 'tmp')
try:
self.use_key = kwargs.pop('use_key')
except KeyError:
self.use_key = False
super(CustomImageField, self).__init__(*args, **kwargs)
def contribute_to_class(self, cls, name):
"""Hook up events so we can access the instance."""
super(CustomImageField, self).contribute_to_class(cls, name)
dispatcher.connect(self._move_image, signal=signals.post_save, sender=cls)
def _move_image(self, instance=None):
"""
Function to move the temporarily uploaded image to a more suitable directory
using the model's get_upload_to() method.
"""
if hasattr(instance, 'get_upload_to'):
src = getattr(instance, self.attname)
if src:
m = re.match(r"%s/(.*)" % self.upload_to, src)
if m:
if self.use_key:
dst = "%s/%d_%s" % (instance.get_upload_to(self.attname), instance.id, m.groups()[0])
else:
dst = "%s/%s" % (instance.get_upload_to(self.attname), m.groups()[0])
basedir = "%s%s/" % (settings.MEDIA_ROOT, os.path.dirname(dst))
mkpath(basedir)
shutil.move("%s%s" % (settings.MEDIA_ROOT, src),"%s%s" % (settings.MEDIA_ROOT, dst))
setattr(instance, self.attname, dst)
instance.save()
def db_type(self):
"""Required by Django for ORM."""
return 'varchar(100)'
class Image(models.Model):
file = CustomImageField(use_key=True, upload_to='tmp')
def get_upload_to(self, attname):
return 'path/to/{0}'.format(self.id)
Just add this function to the model :
def save(self, *args, **kwargs):
if self.pk is None:
saved_image = self.cover_image
self.cover_image = None
super(Item, self).save(*args, **kwargs)
self.cover_image = saved_image
else:
super(Item, self).save(*args, **kwargs)
Updated to the new way of use signals for newer versions of django:
from django.db.models import ImageField, FileField, signals
from django.conf import settings
import shutil, os, glob, re
from distutils.dir_util import mkpath
class CustomImageField(ImageField):
"""Allows model instance to specify upload_to dynamically.
Model class should have a method like:
def get_upload_to(self, attname):
return 'path/to/{0}'.format(self.id)
"""
def __init__(self, *args, **kwargs):
kwargs['upload_to'] = kwargs.get('upload_to', 'tmp')
try:
self.use_key = kwargs.pop('use_key')
except KeyError:
self.use_key = False
super(CustomImageField, self).__init__(*args, **kwargs)
def contribute_to_class(self, cls, name):
"""Hook up events so we can access the instance."""
super(CustomImageField, self).contribute_to_class(cls, name)
signals.post_save.connect(self._move_image, sender=cls)
def _move_image(self, instance, **kwargs):
"""
Function to move the temporarily uploaded image to a more suitable directory
using the model's get_upload_to() method.
"""
if hasattr(instance, 'get_upload_to'):
src = getattr(instance, self.attname)
if src:
m = re.match(r"%s/(.*)" % self.upload_to, str(src))
if m:
if self.use_key:
dst = "%s/%d_%s" % (instance.get_upload_to(self.attname), instance.id, m.groups()[0])
else:
dst = "%s/%s" % (instance.get_upload_to(self.attname), m.groups()[0])
basedir = "%s/%s/" % (settings.MEDIA_ROOT, os.path.dirname(dst))
mkpath(basedir)
shutil.move("%s/%s" % (settings.MEDIA_ROOT, src),"%s/%s" % (settings.MEDIA_ROOT, dst))
setattr(instance, self.attname, dst)
instance.save()
def db_type(self):
"""Required by Django for ORM."""
return 'varchar(100)'
I have a Django project in which I have a view subclassed from the Django CreateView class. This view is used to upload a file to the server, and uses an UploadedFile model which I have created. The UploadedFile also needs to be associated with a project, which is stored as a ForeignKey called project in the UploadedFile model.
The project id is passed in as part of the URL: (r'^projects/(?P<proj_key>\d+)/$', UploadedFileCreateView.as_view(), {}, 'upload-new')
Because project is not really a form field, I know I need to exclude it using a ModelForm; however, even after I have done so, django never enters the form_valid method (if I put a logging call in it, it will never be written to the log, though logging works fine). I'm guessing that the ForeignKey is the culprit because as far as I can tell it worked before I added that in. I don't understand why django doesn't consider the form to be valid even after I excluded project.
Here is my model definition:
class Project(models.Model):
"""This is a project that is owned by a user and contains many UploadedFiles."""
name = models.CharField(max_length=200)
class UploadedFile(models.Model):
"""This represents a file that has been uploaded to the server."""
STATE_UPLOADED = 0
STATE_ANNOTATED = 1
STATE_PROCESSING = 2
STATE_PROCESSED = 4
STATES = (
(STATE_UPLOADED, "Uploaded"),
(STATE_ANNOTATED, "Annotated"),
(STATE_PROCESSING, "Processing"),
(STATE_PROCESSED, "Processed"),
)
status = models.SmallIntegerField(choices=STATES,
default=0, blank=True, null=True)
file = models.FileField(upload_to=settings.XML_ROOT)
project = models.ForeignKey(Project)
def __unicode__(self):
return self.file.name
def name(self):
return os.path.basename(self.file.name)
def save(self, *args, **kwargs):
if not self.status:
self.status = self.STATE_UPLOADED
super(UploadedFile, self).save(*args, **kwargs)
def delete(self, *args, **kwargs):
os.remove(self.file.path)
self.file.delete(False)
super(UploadedFile, self).delete(*args, **kwargs)
class UploadedFileForm(forms.ModelForm):
class Meta:
model = UploadedFile
excludes = ('project',)
Here is my view definition:
class UploadedFileCreateView(CreateView):
model = UploadedFile
def form_valid(self, form):
self.object = form.save(commit=False)
self.object.project_id = self.kwargs['proj_key']
self.object.save()
f = self.request.FILES.get('file')
data = [{'name': f.name,
'url': settings.MEDIA_URL + "files/" + f.name.replace(" ", "_"),
'project': self.object.project.get().pk,
'delete_url': reverse('fileupload:upload-delete',
args=[self.object.id]),
'delete_type': "DELETE"}]
response = JSONResponse(data, {}, response_mimetype(self.request))
response['Content-Disposition'] = 'inline; filename=files.json'
return super(UploadedFileCreateView, self).form_valid(form)
def get_context_data(self, **kwargs):
context = super(UploadedFileCreateView, self).get_context_data(**kwargs)
return context
I see two likely problems:
1) The form keyword is exclude, not excludes.
Generally the recommendation is to favor explicitly listing fields to be included, so you don't accidentally expose any fields you might later add, but exclude will work.
2) You're not actually using your custom form class in the view. Set the form_class attribute to UploadedFileForm.
I have a Django project in which I have a view subclassed from the Django CreateView class. This view is used to upload a file to the server, and uses an UploadedFile model which I have created. The UploadedFile also needs to be associated with a project.
The project id is passed in as part of the URL: (r'^projects/(?P<proj_key>\d+)/$', UploadedFileCreateView.as_view(), {}, 'upload-new')
The problem is that I am not sure where the appropriate place is to associate this key with my model. Is there a method of CreateView or one of its ancestors that I should override that creates the model, or can this be done anywhere in my code in one of the methods I already override (this feels hacky though).
Furthermore, the project attribute of my UploadedFile is defined as a ForeignKey of type Project. How do I get the Project to associate with it?
Here is my model definition:
class Project(models.Model):
"""This is a project that is owned by a user and contains many UploadedFiles."""
name = models.CharField(max_length=200)
class UploadedFile(models.Model):
"""This represents a file that has been uploaded to the server."""
STATE_UPLOADED = 0
STATE_ANNOTATED = 1
STATE_PROCESSING = 2
STATE_PROCESSED = 4
STATES = (
(STATE_UPLOADED, "Uploaded"),
(STATE_ANNOTATED, "Annotated"),
(STATE_PROCESSING, "Processing"),
(STATE_PROCESSED, "Processed"),
)
status = models.SmallIntegerField(choices=STATES,
default=0, blank=True, null=True)
file = models.FileField(upload_to=settings.XML_ROOT)
project = models.ForeignKey(Project)
def __unicode__(self):
return self.file.name
def name(self):
return os.path.basename(self.file.name)
def save(self, *args, **kwargs):
if not self.status:
self.status = self.STATE_UPLOADED
super(UploadedFile, self).save(*args, **kwargs)
def delete(self, *args, **kwargs):
os.remove(self.file.path)
self.file.delete(False)
super(UploadedFile, self).delete(*args, **kwargs)
Here is my view definition:
class UploadedFileCreateView(CreateView):
model = UploadedFile
def form_valid(self, form):
logger.critical("Inside form_valid")
self.object = form.save()
f = self.request.FILES.get('file')
data = [{'name': f.name,
'url': settings.MEDIA_URL + "files/" + f.name.replace(" ", "_"),
'project': self.object.project.get().pk,
'delete_url': reverse('fileupload:upload-delete',
args=[self.object.id]),
'delete_type': "DELETE"}]
response = JSONResponse(data, {}, response_mimetype(self.request))
response['Content-Disposition'] = 'inline; filename=files.json'
return super(UploadedFileCreateView, self).form_valid(form)
def get_context_data(self, **kwargs):
context = super(UploadedFileCreateView, self).get_context_data(**kwargs)
return context
You could do it right where you are calling form.save(). Just pass commit=False so that it won't save it to the db until you add the project id. For example:
self.object = form.save(commit=False)
self.object.project_id = self.kwargs['proj_key']
self.object.save()
Just make sure your form excludes the project field.
EDIT: to exclude the field, add an excludes variable to the form meta class:
class UploadedFileForm(forms.ModelForm):
class Meta:
model = UploadedFile
excludes = ('project',)
When I assign a value to an variable of a Field object, why when I reload the ModelForm isn't reassigned to default?
File
class CustomFile(ImageFile, FieldFile):
def save(self, name, content, save = True):
if self.field.override_name:
self(CustomFile, self).save(self.field.override_name, content, save = save)
else:
self(CustomFile, self).save(generate_name(self.instance, name), content, save = save)
Field
class CustomImageField(ImageField):
attr_class = CustomFile
def __init__(self, overrided_name, *args, **kwargs):
self.overrided_name = overrided_name
super(CustomImageField, self).__init__(*args, **kwargs)
Model
class Test(models.Model):
email = models.EmailField()
file = CustomImageField()
AdminForm
class TestForm(ModelForm):
def __init__(self, *args, **kwargs):
super(TestForm, self).__init__(*args, **kwargs)
self.old_instance = self.instance
Admin
class TestAdmin(Test):
form = TestForm
def save_model(self, request, obj, form, change):
if form.old_instance:
form.old_instance.file.delete(save = True)
form.old_instance.file.field.override_name = form.old_instance.name
obj.save()
admin.site.register(Test, TestAdmin)
My problem is that every image I will upload will have the same name, until I restart the server!..
Why the object doesn't change?! In particular the Field object... when I trace it will result the same object .
I've solved it like so:
class CustomFile(ImageFile, FieldFile):
def save(self, name, content, save = True):
if self.field.override_name:
self(CustomFile, self).save(self.field.override_name, content, save = save)
self.field.override_name = None
else:
self(CustomFile, self).save(generate_name(self.instance, name), content, save = save)
I'm using Django 1.2.6, Python 2.6 and Windows!
I am just guessing, but isnt this part of code wrong?
form = TestForm
Should not it be creating an instance instead of referencing to class?
form = TestForm()
or same thing here?
attr_class = CustomFile - > attr_class = CustomFile()