I have the following model field
logo = models.ImageField(upload_to=municipality_logo_file_name)
and function that uploads the image to the desired location.
def municipality_logo_file_name(instance, filename):
name, extension = os.path.splitext(filename)
return os.path.join('municipality', 'logo', str(uuid.uuid4()) + extension.lower())
In the development mode, a file uploads correctly to for example /media/municipality/logo/1e3cc24a-841b-4577-aa59-b53e5b10420b.png and then displays correctly in the template using
<img src="{{ municipality.logo.url }}" alt="Logo obce" style="max-height: 8rem; width: auto">
In the production, file uploads well, but does not display in the template and the url cannot be followed to display image, the response is
Not Found
The requested resource was not found on this server.
Using IIS and wfastcgi.py.
Any hint as to why this might be happening would be much appreciated.
Try this:
from django.conf import settings
from django.db import models
class Image(models.Model)
image = models.ImageField('Image', upload_to='video-image/')
#property
def image_url(self):
return '%s%s' % (settings.HOST, self.image.url) if self.image else ''
Related
I'm learning flask and i got stuck in this part.
I want to send a image to a img HTML tag without saving it...
here is where i got so far
PYTHON
def serve_pil_image(pil_img):
img_io = BytesIO()
pil_img.save(img_io, 'JPEG', quality=70)
img_io.seek(0)
return send_file(img_io, mimetype='image/jpeg')
#app.route('/upload', methods=["POST"])
def upload():
target = os.path.join(APP_ROOT, 'static/images')
# create image directory if not found
if not os.path.isdir(target):
os.mkdir(target)
# retrieve file from HTML -- NO SAVING
for upload in request.files.getlist("file"):
print('Getting ', upload)
img = Image.open(upload)
return render_template('processing.html')
#app.route('/static/images')
def serve_img():
img = poster
return serve_pil_image(img)
HTML
<img src="{{ url_for('serve_img', filename=img) }}" class="figure-img img-fluid rounded">
If you know a little JavaScript you could create a url within the window using createObjectURL. Store the blob content that got sent by Flask.send_file in the created object. Get a reference to the image tag, and make it point to the in memory 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 have a Django site that I'm creating, and I want some of the pages to have videos embedded in them. These videos aren't part of a model. I just want to be able to use the view to figure out which video file to play, and then pass the file path into the template. All the files are hosted locally (for now, at least).
Is it possible to do with Django? And if so, how do I do it?
There are two ways you can do this -
Method 1: Pass parameter in URL and display video based on that parameter -
If you don't want to use models at any cost, use this, else try method 2.
Assuming you have saved all videos in your media directory and they all have unique names (serving as their ids).
your_app/urls.py -
from django.conf.urls import url
from . import views
urlpatterns = [
url(r'^video/(?P<vid>\w+)/$',views.display_video)
# \w will allow alphanumeric characters or string
]
Add this in the project's settings.py -
#Change this as per your liking
MEDIA_URL = '/media/'
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
your_app/views.py -
from django.conf import settings
from django.shortcuts import render
from django.http import HttpResponse
import os
import fnmatch
def display_video(request,vid=None):
if vid is None:
return HttpResponse("No Video")
#Finding the name of video file with extension, use this if you have different extension of the videos
video_name = ""
for fname in os.listdir(settings.MEDIA_ROOT):
if fnmatch.fnmatch(fname, vid+".*"): #using pattern to find the video file with given id and any extension
video_name = fname
break
'''
If you have all the videos of same extension e.g. mp4, then instead of above code, you can just use -
video_name = vid+".mp4"
'''
#getting full url -
video_url = settings.MEDIA_URL+video_name
return render(request, "video_template.html", {"url":video_url})
Then in your template file, video_template.html, display video as -
<video width="400" controls>
<source src="{{url}}" type="video/mp4">
Your browser does not support HTML5 video.
</video>
Note: There can be performance issue, iterating through all the files in the folder using os.listdir(). Instead, if possible, use a common file extension or use the next method (strongly recommended).
Method 2 : Storing video ids and correspondig file names in database -
Use same settings.py, urls.py and video_template.html as in method 1.
your_app/models.py -
from django.db import models
class videos(models.Model):
video_id = models.CharField(blank=False, max_length=32)
file_name = models.CharField(blank=False, max_length=500)
def __str__(self):
return self.id
your_app/views.py -
from django.conf import settings
from django.shortcuts import render, get_object_or_404
from django.http import HttpResponse
from .models import videos
def display_video(request,vid=None):
if vid is None:
return HttpResponse("No Video")
try:
video_object = get_object_or_404(videos, pk = vid)
except videos.DoesNotExist:
return HttpResponse("Id doesn't exists.")
file_name = video_object.file_name
#getting full url -
video_url = settings.MEDIA_URL+file_name
return render(request, "video_template.html", {"url":video_url})
So if you want to access any page with video id 97veqne0, just goto - localhost:8000/video/97veqne0
I have a Django admin setup where users can upload images. The system only allows for JPEG-formatted images. I put together a validation system to check all the images uploaded are JPEGS. In my Images model I have an override for clean():
class Image(models.Model):
image = models.ImageField(upload_to="images/", blank=True, null=True, help_text='JPEG images only', max_length=100)
...
def clean(self):
import Image
if "images/" in str( self.image ):
i = Image.open( "%s/%s" % ( settings.MEDIA_ROOT, self.image ) )
if i.format != "JPEG":
raise validators.ValidationError, u'You can only upload JPEG images'
The problem is that this will only find an image once it's uploaded and the record is being re-saved. If it's being created for the first time clean() function will be called before the image is saved into the media folder.
Is there a function I could override which will be able to raise an issue with a file being uploaded in the wrong format prior to the record being saved but after the file has at least been stored or is there a way of finding the temporary file location during the execution of clean()?
It's not as fool proof as actually loading up the image with PIL and checking its format, but the field has a name attribute that you can check when cleaning the model.
import re
p = re.compile(r'.*\.(jpg|jpeg)$', re.I)
filename = self.your_file_field.name
if not p.match(filename):
raise ValidationError('You must upload a JPEG image')
in uploaded you can use:
...
import Image
from cStringIO import StringIO
self.image.open()
i = Image.open(StringIO(self.image.file.read())
...
for check if file uploaded:
from django.core.files.uploadedfile import InMemoryUploadedFile
if isinstance(self.image.file, InMemoryUploadedFile):
...
I check all it in django 1.3
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