I'm looking to the best way to overriding the _get_url method from ImageField, I need to customize the url since I don't want the default returned url (I distribute this image trhough a view to manage ACL on it so the url based on the MEDIA_ROOT is wrong).
Should I have to create my own ImageField ? or is there a solution using less code ?
Thanks in advance
Fabien
The url returned be _get_url is actually generated by the used Storage class; it would probably make more sense to create your own Storage and use it when creating the ImageField!
See: http://docs.djangoproject.com/en/dev/topics/files/#file-storage, http://docs.djangoproject.com/en/dev/howto/custom-file-storage/
You will neet to override the Storage.url method for that!
Thanks to lazerscience, here my final solution :
from django.core.files.storage import FileSystemStorage
from django.db.models import get_model
from django.core.urlresolvers import reverse
from django.db import models
from django.conf import settings
class PhotographerStorage(FileSystemStorage):
def __init__(self, location=None):
super(PhotographerStorage, self).__init__(location)
def url(self, name):
photo_model = get_model('photographer', 'photo')
photo = photo_model.objects.get(original_image=name)
url = super(PhotographerStorage, self).url(name)
return '%s?size=%d' % (reverse('photographer_photo_display',
args=[photo.slug]), 300)
fs = PhotographerStorage(location=settings.PHOTOGRAPHER_LOCATION)
class Photo(models.Model):
...
original_image = models.ImageField(storage=fs)
...
It works like a charm :)
Related
I am trying to add a dynamic instead of a hard coded get_inspect_url method to a PageModel:
class MyModel(Page):
# ...
# this is what I have now, which breaks in prod due to a different admin url
def get_inspect_url(self):
inspect_url = f"/admin/myapp/mymodel/inspect/{self.id}/"
return inspect_url
I know that you can use the ModelAdmin.url_helper_class in wagtail_hooks.py to get admin urls for an object, but I can't figure a way to do the same on the model level as a class method. Is there a way?
thanks to #Andrey Maslov tip I found a way:
from django.urls import reverse
def get_inspect_url(self):
return reverse(
f"{self._meta.app_label}_{self._meta.model_name}_modeladmin_inspect",
args=[self.id],
)
The basic form of a Wagtail admin url is:
[app_label]_[model_name]_modeladmin_[action]
Just change the [action] to one of the following to generated other instance level admin urls:
edit
delete
if you have in you urls.py line like this
path(MyModel/view/<id:int>/', YourViewName, name='mymodel-info'),
then you can add to your models.py this lines
from django.urls import reverse
...
class MyModel(Page):
def get_inspect_url(self):
inspect_url = reverse('mymodel-info', kwargs={'id': self.id})
return inspect_url
I am using django-rest-framework in conjuntion with django-ckeditor. I'm serving some images with absolute url-s without any problem. But images and files uploaded by ckeditor are served as relative paths, and they can't be displayed client side since it is in a different domain.
Here is an example of what I'm getting:
{
image: "http://example.com/media/myimage.png",
body: "<p>download my file</p>"
}
And this is what I woul like to get:
{
image: "http://example.com/media/myimage.png",
body: "<p>download my file</p>"
}
Edit:
This would be the model of my example:
from django.db import models
from ckeditor_uploader.fields import RichTextUploadingField
image: models.ImageField()
body: RichTextUploadingField(blank=True,null=True)
I would use a custom serializer to fix that:
from rest_framework import serializers
def relative_to_absolute(url):
return 'http://127.0.0.1:8000' + url
class FileFieldSerializer(serializers.Field):
def to_representation(self, value):
url = value.url
if url and url.startswith('/'):
url = relative_to_absolute(url)
return url
When filefield.url contains a relative url, relative_to_absolute() is called to prepend the domain.
Here I just used a constant string; you can either save it in your settings, or, if Django Site framework is installed, retrieve it as follows:
from django.contrib.sites.models import Site
domain=Site.objects.get_current().domain
Sample usage of the custom serializer:
class Picture(BaseModel):
...
image = models.ImageField(_('Image'), null=True, blank=True)
...
class PictureSerializer(serializers.ModelSerializer):
image = FileFieldSerializer()
class Meta:
model = Picture
fields = '__all__'
Variation for RichTextUploadingField
If, on the other hand, you're using RichTextUploadingField by CKEditor, your field is, basically, a TextField where an HTML fragment is saved upon images
upload.
In this HTML fragment, CKEditor will reference the uploaded images with a relative path, for very good reasons:
your site will still work if the domain is changed
the development instance will work in localhost
after all, we're using Django, not WordPress ;)
So, I wouldn't touch it, and fix the path at runtime in a custom serializer instead:
SEARCH_PATTERN = 'href=\\"/media/ckeditor/'
SITE_DOMAIN = "http://127.0.0.1:8000"
REPLACE_WITH = 'href=\\"%s/media/ckeditor/' % SITE_DOMAIN
class FixAbsolutePathSerializer(serializers.Field):
def to_representation(self, value):
text = value.replace(SEARCH_PATTERN, REPLACE_WITH)
return text
Alternatively, domain can be saved in settings:
from django.conf import settings
REPLACE_WITH = 'href=\\"%s/media/ckeditor/' % settings.SITE_DOMAIN
or retrieved from Django Site framework as follows:
from django.contrib.sites.models import Site
REPLACE_WITH = 'href=\\"{scheme}{domain}/media/ckeditor/'.format(
scheme="http://",
domain=Site.objects.get_current().domain
)
You might need to adjust SEARCH_PATTERN according to your CKEditor configuration; the more specific, the better.
Sample usage:
class Picture(BaseModel):
...
body = RichTextUploadingField(blank=True,null=True)
...
class PictureSerializer(serializers.ModelSerializer):
body = FixAbsolutePathSerializer()
class Meta:
model = Picture
fields = '__all__'
I've created a Configuration model in django so that the site admin can change some settings on the fly, however some of the models are reliant on these configurations. I'm using Django 2.0.2 and Python 3.6.4.
I created a config.py file in the same directory as models.py.
Let me paracode (paraphase the code? Real Enum has many more options):
# models.py
from .config import *
class Configuration(models.Model):
starting_money = models.IntegerField(default=1000)
class Person(models.Model):
funds = models.IntegarField(default=getConfig(ConfigData.STARTING_MONEY))
# config.py
from .models import Configuration
class ConfigData(Enum):
STARTING_MONEY = 1
def getConfig(data):
if not isinstance(data, ConfigData):
raise TypeError(f"{data} is not a valid configuration type")
try:
config = Configuration.objects.get_or_create()
except Configuration.MultipleObjectsReturned:
# Cleans database in case multiple configurations exist.
Configuration.objects.exclude(Configuration.objects.first()).delete()
return getConfig(data)
if data is ConfigData.MAXIMUM_STAKE:
return config.max_stake
How can I do this without an import error? I've tried absolute imports
You can postpone loading the models.py by loading it in the getConfig(data) function, as a result we no longer need models.py at the time we load config.py:
# config.py (no import in the head)
class ConfigData(Enum):
STARTING_MONEY = 1
def getConfig(data):
from .models import Configuration
if not isinstance(data, ConfigData):
raise TypeError(f"{data} is not a valid configuration type")
try:
config = Configuration.objects.get_or_create()
except Configuration.MultipleObjectsReturned:
# Cleans database in case multiple configurations exist.
Configuration.objects.exclude(Configuration.objects.first()).delete()
return getConfig(data)
if data is ConfigData.MAXIMUM_STAKE:
return config.max_stake
We thus do not load models.py in the config.py. We only check if it is loaded (and load it if not) when we actually execute the getConfig function, which is later in the process.
Willem Van Onsem's solution is a good one. I have a different approach which I have used for circular model dependencies using django's Applications registry. I post it here as an alternate solution, in part because I'd like feedback from more experienced python coders as to whether or not there are problems with this approach.
In a utility module, define the following method:
from django.apps import apps as django_apps
def model_by_name(app_name, model_name):
return django_apps.get_app_config(app_name).get_model(model_name)
Then in your getConfig, omit the import and replace the line
config = Configuration.objects.get_or_create()
with the following:
config_class = model_by_name(APP_NAME, 'Configuration')
config = config_class.objects.get_or_create()
Is it possible to get object with comments related to it? Right now django comment framework creates query for every object which has related comments and another queries for comments owners. Can I somehow avoid this? I use django 1.4 so prefetch_related is allowed.
You could create a function that caches the count:
from django.contrib.contenttypes.models import ContentType
from django.contrib import comments
def get_comment_count_key(model):
content_type = ContentType.objects.get_for_model(model)
return 'comment_count_%s_%s' % (content_type.pk, model.pk)
def get_comment_count(model):
key = get_comment_count_key(model)
value = cache.get(key)
if value is None:
value = comments.get_model().objects.filter(
content_type = ContentType.objects.get_for_model(model),
object_pk = model.pk,
site__pk = settings.SITE_ID
).count()
cache.set(key, value)
return value
You could extend the Comment model and add get_comment_count there. Or put get_comment_count as a template filter. It doesn't matter.
Of course, you would also need cache invalidation when a new comment is posted:
from django.db.models import signals
from django.contrib import comments
def refresh_comment_count(sender, instance, **kwargs):
cache.delete(get_comment_count_key(instance.content_object))
get_comment_count(instance.content_object)
post_save.connect(refresh_comment_count, sender=comments.get_model())
post_delete.connect(refresh_comment_count, sender=comments.get_model())
You could improve this last snippet, by using cache.incr() on comment_was_posted, and cache.decr() on post_delete but that's left as an exercise for you :)
I am having error after error trying to upload and resize images to s3 with pil and botos3 and the django default_storage. I am trying to do this on save in the admin.
here is the code:
from django.db import models
from django.forms import CheckboxSelectMultiple
import tempfile
from django.conf import settings
from django.core.files.base import ContentFile
from django.core.files.storage import default_storage as s3_storage
from django.core.cache import cache
from datetime import datetime
import Image, os
import PIL.Image as PIL
import re, os, sys, urlparse
class screenshot(models.Model):
title = models.CharField(max_length=200)
slug = models.SlugField(max_length=200)
image = models.ImageField(upload_to='screenshots')
thumbnail = models.ImageField(upload_to='screenshots-thumbs', blank=True, null=True, editable=False)
def save(self):
super(screenshot, self).save() # Call the "real" save() method
if self.image:
thumb = Image.open(self.image.path)
thumb.thumbnail(100, 100)
filename = str(self.slug)
temp_image = open(os.path.join('tmp',filename), 'w')
thumb.save(temp_image, 'JPEG')
from django.core.files import File
thumb_data = open(os.path.join('/tmp',filename), 'r')
thumb_file = File(thumb_data)
new_file.thumb.save(str(self.slug) + '.jpg', thumb_file)
def __str__(self):
return self.title
This is just one of the many ways I have tried to get it working, and I either get (2, 'No such file or directory') or some other error.
Please can someone help me to get it working. I want it to use the django backend to get the image uploaded to be resized and saved as the thumbnail and then saved. Let me know if you need to know any information. I would be happy to use the django snippet - http://djangosnippets.org/snippets/224/ but I don't know what data to feed it. I get the same IOErrors and 'no such path/filename' even though the main image is uploading to s3 fine. I have also tried things like:
myimage = open(settings.MEDIA_URL + str(self.image))
myimage_io = StringIO.StringIO()
imageresize = myimage.resize((100,100), Image.ANTIALIAS)
imageresize.save('resize_100_100_aa.jpg', 'JPEG', quality=75)
It's been 3 days of looking now so I am starting to go spare! Thanks
I had a similar problem, but in my case using sorl-thumbnail was not an option. I found that I can open an Image directly from S3BotoStorage by passing in a file descriptor instead of a path.
So instead of
thumb = Image.open(self.image.path)
use
thumb = Image.open(s3_storage.open(self.image.name))
Then you can process and save the new file locally as you were doing before.
Why don't you try sorl-thumbnail. It has the exact same interface as the default ImageField django provides and it seems like it would be a lot nicer to work with than the roll-your-own support.
Storage support
Pluggable Engine support (PIL, pgmagick)
Pluggable Key Value Store support (redis, cached db)
Pluggable Backend support
Admin integration with possibility to delete
Dummy generation
Flexible, simple syntax, generates no html
ImageField for model that deletes thumbnails
CSS style cropping options
Margin calculation for vertical positioning