I am using django-extensions.AutoSlugField and django.db.models.ImageField.
To customize image name uploaded for django.db.models.ImageField, what I did:
from django.utils.text import slugify
# idea is to make image name the same as the automatically generated slug, however don't work
def update_image_name(instance, filename):
# this debug instance, and instance.slug is empty string
print(instance.__dict__)
# I attempt to use slugify directly and see that it's not the same as the output generated by AutoSlugField
# E.g. If I create display_name "shawn" 2nd time, AutoSlugField will return "shawn-2", but slugify(display_name) return "shawn"
print(slugify(instance.display_name))
return f"images/{instance.slug}.jpg"
class Object(models.Model):
...
display_name = models.TextField()
...
# to customize uploaded image name
image = models.ImageField(blank=True, upload_to=update_image_name)
...
# create slug automatically from display_name
slug = AutoSlugField(blank=True, populate_from=["display_name"]
Based on what I debug, when I call instance inside update_image_name, slug is empty string.
If I understand correctly slug is only created at event save, so when I call ImageField instance, slug is not yet created, therefore empty string.
I think it might have something to do with event post save. However, I am not sure if that's the real reason or how to do that.
How can I get the automatically generated slug as my customized image name?
That's a tricky one because the order the fields are getting saved matters.
The brute-force attack I'm suggesting would be to override the save method of your model and manually call the create_slug method before everything else ensuring the slug is set:
from django.utils.encoding import force_str
[...]
class Object(models.Model):
[...]
def save(self, *args, **kwargs):
self.slug = force_str(self._meta.get_field('slug').create_slug(self, False))
super(Object, self).save(*args, **kwargs)
That's what AutoSlugField does, refer to the code here. self._meta.get_field('slug') get's the slug field definition and then we just call the create_slug method.
Tested under Python 3.7.9 & Django 3.1.5 like this:
from django.core.files.uploadedfile import SimpleUploadedFile
o = Object()
o.display_name = "foo bar"
o.image = SimpleUploadedFile(name='test_image.png', content=open('/path/to/test/image.png', 'rb').read(), content_type='image/png')
o.save()
Then I see update_image_name return images/foo-bar.jpg.
Related
I know that they don't do this, but for one of my pet-projects I want a strange thing: store jinja-templates in the database (and be able to edit them through the admin panel).
There is something like this model (in models.py):
class TbTemplate(models.Model):
szFileName = models.CharField(
primary_key=True,
db_index=True,
unique=True,
verbose_name="Path/Name"
)
szJinjaCode = models.TextField(
verbose_name='Template',
help_text='Template Code (jinja2)'
)
szDescription = models.CharField(
max_length=100,
verbose_name='Description'
)
def __unicode__(self):
return f"{self.szFileName} ({self.szDescription})"
def __str__(self):
return self.__unicode__()
class Meta:
verbose_name = '[…Template]'
verbose_name_plural = '[…Templates]'
Next, in view.py you can do something like this:
# -*- coding: utf-8 -*-
from django.http import HttpRequest, HttpResponse
from django.template.loader import render_to_string
from web.models import TbTemplate
def something(request: HttpRequest, template: str) -> HttpResponse:
"""
:param request: http-in
:param template: Template name
:return response: http-out
"""
to_template = {}
# ...
# ... do smth
# ...
tmpl = TbTemplate.objects.get(pk=template)
html = render_to_string(tmpl.szJinjaCode, to_template)
return HttpResponse(html)
And everything works. Templates to available for editing through the admin panel (of course, you need to hang a "mirror"-like widget for syntax highlighting, etc.)...
But I want to use in jinja templates like: {% include "something_template.jinja2" %} ... And for this it is necessary that the templates are not only in the database, but also stored as files in the templates-folder.
In addition, templates are easier to create and edit in IDEs, and access to templates through the admin panel only for cosmetic changes.
And then I need to somehow intercept the "read" method in/for the admin panel. So that if a template in the TbTemplate table is opened for editing in the admin panel, then for the szJinjaCode it was read not from the database, but from the corresponding szFileName-file.
How to do this?
It is done in two steps:
Firstly,
in models.py we will override the save() method for the TbTemplate model. At the same time, we can override delete() method, so that it do not delete anything (or vice versa, it deletes not only the entry in the database, but also the corresponding teplate-file... or deletes the entry in the database, and renames the corresponding file...). We get this model:
# -*- coding: utf-8 -*-
from django.db import models
from project.settings import *
import os
class TbTemplate(models.Model):
szFileName = models.CharField(
primary_key=True, db_index=True, unique=True,
verbose_name="Path/Name"
)
szJinjaCode = models.TextField(
default='', null=True, blank=True,
verbose_name='Template',
help_text='Template Code (jinja2)'
)
szDescription = models.CharField(
max_length=100,
verbose_name='Description'
)
def __unicode__(self):
return f"{self.szFileName} ({self.szDescription})"
def __str__(self):
return self.__unicode__()
def save(self, *args, **kwargs):
path_filename = TEMPLATES_DIR / self.szFileName
if not os.path.exists(os.path.dirname(path_filename)):
os.makedirs(os.path.dirname(path_filename))
with open(path_filename, "w+", encoding="utf-8") as file:
file.write(self.szJinjaCode)
# TODO: for production, need to add some code for modify
# touch_reload file for uWSGI reload
super(TbTemplate, self).save(*args, **kwargs)
def delete(self, *args, **kwargs):
pass
# ... or do smth ... and after:
# super(TbTemplate, self).delete(*args, **kwargs)
class Meta:
verbose_name = '[…Template]'
verbose_name_plural = '[…Templates]'
now, when changing and creating a template through Django-Admin, a corresponding template file will be create/modify.
Secondly,
in the admin.py file, when define the admin.ModelAdmin class for the TbTemplate control model, it is necessary to override the get_fields () method, which is responsible for getting the fields in the admin form. (I had to look for a method stupidly trying many cases that are made something similar, but not that). As a result, something like this admin.py:
# -*- coding: utf-8 -*-
from django.contrib import admin
from my_appweb.models import TbTemplate
from project.settings import *
class AdminTemplate(admin.ModelAdmin):
list_display = ('szFileName', 'szDescription')
list_display_links = ('szFileName', 'szDescription', )
def get_fields(self, request, obj=None):
try:
with open(Path(TEMPLATES_DIR) / obj.szFileName, "r", encoding="utf-8") as file:
obj.szJinjaCode = file.read()
except (AttributeError, FileNotFoundError, TypeError):
pass
return ['szFileName', 'szDescription', 'szJinjaCode']
admin.site.register(TbTemplate, AdminTemplate)
Now, if some "external forces" change the template file (or someone changes the template code in the database bypassing the admin panel), then when you open the template for editing in the admin panel, the data will still be received from the file.
It's all
P.S. in the settnig.py of the project, you need to add something like:
TEMPLATES_DIR = BASE_DIR / 'templates-jinja2'
So that the model and the admin panel know in which directory to pour the template files.
Interesting question.
The first part - inclusion tag
Your foundation - you want to use inclusion tag. Therefore you want to save something in file. But you can simply override template loader, who get before file the template from the database:
#settings.py:
TEMPLATES = [
{
'BACKEND': 'myapp.backends.MyTemplate',
... # other staff
},
]
in myapp/backends.py:
class MyTemplate(Jinja2):
def get_template(self, template_name):
template = TbTemplate.objects.filter(pk=template_name).first()
if template:
return self.from_string(template.szJinjaCode)
return super().get_template(template_name)
After that - every template can be saved in DB, {% include %} call template_backend which get template from database before file-template.
The second part. Save template to file/database.
If you do it, you don't need the first part, every time the template-file should be saved/changed.
class TbTemplateAdmin(ModelAdmin):
def save_model(self, request, obj, *args, **kwargs):
super().save(request, obj, *args, **kwargs)
with open( Path(path_to_templates) / obj.szFileName, "w+" ) as template:
template.write(obj.szJinjaCode)
The Third part - get the file in admin on get_object:
class TbTemplateAdmin(ModelAdmin):
def get_object(self, *args, **kwargs):
obj = super().get_object(self, *args, **kwargs)
with open( Path(path_to_templates) / obj.szFileName, "r" ) as template:
obj.szJinjaCode = template.read()
return obj
The Last part - convert the new file templates to objects:
In our projects we add automatically the new templates to database, spoiler - with inclusion tags. In your case - you can create an ModelAdmin.action to add templates in database. I don't solve it for you, try to do something yourself. I hope for your understanding
Only one fing I lost here. If you use Cached Template Loader, and you should use it on production, in this case you should refresh cache for changed templates. Don't forget about it.
I'm restricting the upload button to allow only csv files.
I need help please to append _hello at the end of each file uploaded by the user, but before the extension. (e.g. user_file_name.csv becomes automatically user_file_name_hello.csv)
Optional: I'd like the original file to be first renamed automatically, then saved to my uploads directory.
models.py
from django.db import models
# validation method to check if file is csv
from django.core.exceptions import ValidationError
def validate_file_extension(value):
if not value.name.endswith('.csv'):
raise ValidationError(u'Only CSV files allowed.')
# Create your models here.
class user_file(models.Model):
user_file_csv = models.FileField(upload_to='documents/user_files/', validators=[validate_file_extension])
forms.py
from django import forms
from .models import user_file
from django.forms import FileInput
class user_file_form(forms.ModelForm):
class Meta:
model = user_file
widgets = {'user_file_csv': FileInput(attrs={'accept': 'text/csv'})}
fields = ('user_file_csv',)
Thank you!
Maybe you need something like this:
class FileUploadUtil:
#staticmethod
def my_files_path(instance, filename):
name, file_extention = os.path.splitext(filename)
name = 'prefix-{}-{}-sufix.{}'.format(name, instance.id, file_extention)
return "my_files/{}".format(name)
class MyModel(models.Model):
# Other fields
# ...
my_file = models.FileField(max_length=300, upload_to=FileUploadUtil.my_files_path)
Optional: I'd like the original file to be first renamed automatically, then saved to my uploads directory.
You can override save() method. Check here
Django document
Maybe You need decorator.
from pathlib import Path
def rename_helper(path: str, append_text: str):
stem, suffix = Path(path).stem, Path(path).suffix
return f"{stem}{append_text}{suffix}"
def rename_previous_image(func):
""" return wrapper object """
def wrapper(*args, **kwargs):
self = args[0]
model = type(self)
previous_obj = model.objects.filter(pk=self.pk)
if previous_obj.exists():
old_name_with_path = Path(str(previous_obj[0].user_file_csv))
Path.rename(old_name_with_path , rename_helper(path=old_name_with_path , append_text="_hello"))
return func(*args, **kwargs)
return wrapper
And, You can decorate your model save() method.
class MyModel(models.Model):
# Other fields
# ...
my_file = models.FileField(max_length=300, upload_to=FileUploadUtil.my_files_path)
#rename_previous_image
def save(self, **kwargs):
super(user_file, self).save(**kwargs) # You must add This row.
besides,
recommend rename your user_file class
like UserFile
Check This PEP 8
Have a good day.
Can anyone see any issues with the code below? It's my save function for a model it gives them a GUID on first save. My my problem is when I save a new recipient (in the admin) it overwrites the last one added. Updates seem to work perfectly tho.
part of Models.py
class GUID():
make = hashlib.sha1(str(random.random())).hexdigest()
def save(self, *args, **kwargs):
if not self.recipientid:
self.recipientid = GUID.make
super(Recipient, self).save(*args, **kwargs)
GUID.make will be set at the time the GUID class is created, it won't re-calculated each time it's run. I don't know the rest of the context of how you're using GUID, but I'd have it be a function:
class GUID(object):
#staticmethod
def make():
return hashlib.sha1(str(random.random())).hexdigest()
...
def save(self, *args, **kwargs):
if not self.recipientid:
self.recipientid = GUID.make()
super(Recipient, self).save(*args, **kwargs)
Generally speaking, the way to do what you're trying to do is with a default lambda (in this example using a standard python uuid):
from django.db import models
from uuid import uuid4
class YourModel(models.Model):
# ...
recipientid = models.CharField(max_length=32, default=lambda: uuid4().hex)
I have a Django model with an ImageField() field. Now I'd like to rename the filename of the image (based on a unique CharField of the same model) before it gets saved to the filesystem. Additionally, if an image with the same filename already exists, the existing file should be renamed and the newly uploaded file should keep its filename.
I am not quite sure what's the correct or preferred way to do it. Should I override ModelAdmin.save_model(), do it in the Model.save() method or write an own custom file storage?
Can anyone give me some hits how I can accomplish this? Any tips or sample code are greatly appreciated.
You can combine two mechanisms here: passing an upload_to argument in the field's definition and a custom FileSystemStorage backend.
Here is a dummy model with an upload_to callable:
def upload_bar(instance, filename):
# Do something with the filename
return new_filename
class Foo(models.Model):
bar = models.ImageField(upload_to=upload_bar)
...
And a custom dummy FileSystemStorage backend:
from django.core.files.storage import FileSystemStorage
class OverwriteStorage(FileSystemStorage):
def _save(self, name, content):
if self.exists(name):
# Rename it
return super(OverwriteStorage, self)._save(name, content)
def get_available_name(self, name):
return name
However I would be very cautious in meddling with existing files (ie changing existing files' names). Note that Django does not remove the file from the filesystem even when the object is deleted.
Here's the solution to get unique filenames like 18f6ad9f-5cec-4420-abfd-278bed78ee4a.png
models.py
import os
import uuid
from django.conf import settings
from django.db import models
def make_upload_path(instance, filename):
file_root, file_ext = os.path.splitext(filename)
dir_name = '{module}/{model}'.format(module=instance._meta.app_label, model=instance._meta.module_name)
file_root = unicode(uuid.uuid4())
name = os.path.join(settings.MEDIA_ROOT, dir_name, file_root + file_ext.lower())
# Delete existing file to overwrite it later
if instance.pk:
while os.path.exists(name):
os.remove(name)
return os.path.join(dir_name, file_root + file_ext.lower())
class YourModel(models.Model):
title = models.CharField(max_length=100)
image = models.ImageField(blank=True, upload_to=make_upload_path)
you can access self.image.name from Model.save()
http://lightbird.net/dbe/photo.html
I am using 1.2.5 with a standard ImageField and using the built-in storage backend. Files upload fine but when I remove an entry from admin the actual file on the server does not delete.
You can receive the pre_delete or post_delete signal (see #toto_tico's comment below) and call the delete() method on the FileField object, thus (in models.py):
class MyModel(models.Model):
file = models.FileField()
...
# Receive the pre_delete signal and delete the file associated with the model instance.
from django.db.models.signals import pre_delete
from django.dispatch.dispatcher import receiver
#receiver(pre_delete, sender=MyModel)
def mymodel_delete(sender, instance, **kwargs):
# Pass false so FileField doesn't save the model.
instance.file.delete(False)
Try django-cleanup
pip install django-cleanup
settings.py
INSTALLED_APPS = (
...
'django_cleanup.apps.CleanupConfig',
)
Django 1.5 solution: I use post_delete for various reasons that are internal to my app.
from django.db.models.signals import post_delete
from django.dispatch import receiver
#receiver(post_delete, sender=Photo)
def photo_post_delete_handler(sender, **kwargs):
photo = kwargs['instance']
storage, path = photo.original_image.storage, photo.original_image.path
storage.delete(path)
I stuck this at the bottom of the models.py file.
the original_image field is the ImageField in my Photo model.
This code runs well on Django 1.4 also with the Admin panel.
class ImageModel(models.Model):
image = ImageField(...)
def delete(self, *args, **kwargs):
# You have to prepare what you need before delete the model
storage, path = self.image.storage, self.image.path
# Delete the model before the file
super(ImageModel, self).delete(*args, **kwargs)
# Delete the file after the model
storage.delete(path)
It's important to get the storage and the path before delete the model or the latter will persist void also if deleted.
You need to remove the actual file on both delete and update.
from django.db import models
class MyImageModel(models.Model):
image = models.ImageField(upload_to='images')
def remove_on_image_update(self):
try:
# is the object in the database yet?
obj = MyImageModel.objects.get(id=self.id)
except MyImageModel.DoesNotExist:
# object is not in db, nothing to worry about
return
# is the save due to an update of the actual image file?
if obj.image and self.image and obj.image != self.image:
# delete the old image file from the storage in favor of the new file
obj.image.delete()
def delete(self, *args, **kwargs):
# object is being removed from db, remove the file from storage first
self.image.delete()
return super(MyImageModel, self).delete(*args, **kwargs)
def save(self, *args, **kwargs):
# object is possibly being updated, if so, clean up.
self.remove_on_image_update()
return super(MyImageModel, self).save(*args, **kwargs)
You may consider using a pre_delete or post_delete signal:
https://docs.djangoproject.com/en/dev/topics/signals/
Of course, the same reasons that FileField automatic deletion was removed also apply here. If you delete a file that is referenced somewhere else you will have problems.
In my case this seemed appropriate because I had a dedicated File model to manage all of my files.
Note: For some reason post_delete doesn't seem to work right. The file got deleted, but the database record stayed, which is completely the opposite of what I would expect, even under error conditions. pre_delete works fine though.
Maybe it's a little late. But the easiest way for me is to use a post_save signal. Just to remember that signals are excecuted even during a QuerySet delete process, but the [model].delete() method is not excecuted during the QuerySet delete process, so it's not the best option to override it.
core/models.py:
from django.db import models
from django.db.models.signals import post_delete
from core.signals import delete_image_slide
SLIDE1_IMGS = 'slide1_imgs/'
class Slide1(models.Model):
title = models.CharField(max_length = 200)
description = models.CharField(max_length = 200)
image = models.ImageField(upload_to = SLIDE1_IMGS, null = True, blank = True)
video_embed = models.TextField(null = True, blank = True)
enabled = models.BooleanField(default = True)
"""---------------------------- SLIDE 1 -------------------------------------"""
post_delete.connect(delete_image_slide, Slide1)
"""--------------------------------------------------------------------------"""
core/signals.py
import os
def delete_image_slide(sender, **kwargs):
slide = kwargs.get('instance')
try:
os.remove(slide.image.path)
except:
pass
This functionality will be removed in Django 1.3 so I wouldn't rely on it.
You could override the delete method of the model in question to delete the file before removing the entry from the database completely.
Edit:
Here is a quick example.
class MyModel(models.Model):
self.somefile = models.FileField(...)
def delete(self, *args, **kwargs):
somefile.delete()
super(MyModel, self).delete(*args, **kwargs)
Using the post_delete is for sure the right way to go. Sometimes though things can go wrong, and files don't get deleted. There is of course the case that you have a bunch of old files that weren't deleted before post_delete was used. I created a function that deletes files for objects based on if the file the object references does not exist then delete object, if the file does not have an object, then also delete, also it can delete based on an "active" flag for an object.. Something I added to most of my models. You have to pass it the objects you want to check, the path to the objects files, the file field and a flag to delete inactive objects:
def cleanup_model_objects(m_objects, model_path, file_field='image', clear_inactive=False):
# PART 1 ------------------------- INVALID OBJECTS
#Creates photo_file list based on photo path, takes all files there
model_path_list = os.listdir(model_path)
#Gets photo image path for each photo object
model_files = list()
invalid_files = list()
valid_files = list()
for obj in m_objects:
exec("f = ntpath.basename(obj." + file_field + ".path)") # select the appropriate file/image field
model_files.append(f) # Checks for valid and invalid objects (using file path)
if f not in model_path_list:
invalid_files.append(f)
obj.delete()
else:
valid_files.append(f)
print "Total objects", len(model_files)
print "Valid objects:", len(valid_files)
print "Objects without file deleted:", len(invalid_files)
# PART 2 ------------------------- INVALID FILES
print "Files in model file path:", len(model_path_list)
#Checks for valid and invalid files
invalid_files = list()
valid_files = list()
for f in model_path_list:
if f not in model_files:
invalid_files.append(f)
else:
valid_files.append(f)
print "Valid files:", len(valid_files)
print "Files without model object to delete:", len(invalid_files)
for f in invalid_files:
os.unlink(os.path.join(model_path, f))
# PART 3 ------------------------- INACTIVE PHOTOS
if clear_inactive:
#inactive_photos = Photo.objects.filter(active=False)
inactive_objects = m_objects.filter(active=False)
print "Inactive Objects to Delete:", inactive_objects.count()
for obj in inactive_objects:
obj.delete()
print "Done cleaning model."
This is how you can use this:
photos = Photo.objects.all()
photos_path, tail = ntpath.split(photos[0].image.path) # Gets dir of photos path, this may be different for you
print "Photos -------------->"
cleanup_model_objects(photos, photos_path, file_field='image', clear_inactive=False) # image file is default
make sure you write "self" before the file. so example above should be
def delete(self, *args, **kwargs):
self.somefile.delete()
super(MyModel, self).delete(*args, **kwargs)
I've forgotten the "self" before my file and that didn't work as it was looking in the global namespace.
If you already have number of unused files in your project and want to delete them, you can use django utility django-unused-media
Django 2.x Solution:
There's no need to install any packages! It's very easy to handle in Django 2. I've tried following solution using Django 2 and SFTP Storage (however I think it would work with any storages)
First write a Custom Manager. So if you want to be able to delete files of a model by using objects methods, you must write and use a [Custom Manager][3] (for overriding delete() method of objects):
class CustomManager(models.Manager):
def delete(self):
for obj in self.get_queryset():
obj.delete()
Now you must delete image before deleting deleting the model itself and for assigning the CustomManager to the model, you must initial objects inside your model:
class MyModel(models.Model):
image = models.ImageField(upload_to='/pictures/', blank=True)
objects = CustomManager() # add CustomManager to model
def delete(self, using=None, keep_parents=False):
objects = CustomManager() # just add this line of code inside of your model
def delete(self, using=None, keep_parents=False):
self.image.storage.delete(self.song.name)
super().delete()
I may have a special case since I am using the upload_to option on my file field with dynamic directory names but the solution I found was to use os.rmdir.
In models:
import os
...
class Some_Model(models.Model):
save_path = models.CharField(max_length=50)
...
def delete(self, *args,**kwargs):
os.rmdir(os.path.join(settings.MEDIA_ROOT, self.save_path)
super(Some_Model,self).delete(*args, **kwargs)