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
Related
I have a django-cms site, that uses i18n_patterns in urls.py. This works, urls are built like /lang/here-starts-the-normal/etc/.
Now, I would like to have urls like this: /prefix-lang/here-starts.... As there will be a couple of country specific domains, this wille be like /ch-de/here-... for Switzerland/.ch domain, /us-en/here-starts.... for the states, and some more. So, when the url would be /ch-de/..., the LANGUAGE would still be de. Hope this is clear?
As the content is filled with existing LANGUAGES=(('de', 'DE'), ('en', 'EN'), ...), I cannot change LANGUAGES for every domain - no content would be found in the cms, modeltranslation, only to mention those two.
How can I prefix the language slug in i18n_patterns? Is it possible at all?
I think a way without hacking Django too much would be to use URL rewrite facility provided by the webserver you run, for example, for mod_wsgi you can use mod_rewrite, similar facility exists also for uWSGI.
You may need to also post-process the output from Django to make sure that any links are also correctly re-written to follow the new schema. Not the cleanest approach but seems doable.
Working example, though the country/language order is reversed (en-ch instead of ch-en), to have it as django expects it when trying to find a language (ie, setting language to "en-ch", it will find "en", if available).
This solution involves a modified LocaleMiddleware, i18n_patterns, LocaleRegexResolver. It supports no country, or a 2 char country code, setup with settings.SITE_COUNTRY. It works by changing urls to a lang-country mode, but the found language code in middleware will still be language only, 2 chars, and work perfectly with existing LANGUAGES, that contain 2 chars language codes.
custom_i18n_patterns.py - this just uses our new resolver, see below
from django.conf import settings
from ceco.resolvers import CountryLocaleRegexURLResolver
def country_i18n_patterns(*urls, **kwargs):
"""
Adds the language code prefix to every URL pattern within this
function. This may only be used in the root URLconf, not in an included
URLconf.
"""
if not settings.USE_I18N:
return list(urls)
prefix_default_language = kwargs.pop('prefix_default_language', True)
assert not kwargs, 'Unexpected kwargs for i18n_patterns(): %s' % kwargs
return [CountryLocaleRegexURLResolver(list(urls), prefix_default_language=prefix_default_language)]
resolvers.py
import re
from django.conf import settings
from django.urls import LocaleRegexURLResolver
from modeltranslation.utils import get_language
class CountryLocaleRegexURLResolver(LocaleRegexURLResolver):
"""
A URL resolver that always matches the active language code as URL prefix.
extended, to support custom country postfixes as well.
"""
#property
def regex(self):
language_code = get_language() or settings.LANGUAGE_CODE
if language_code not in self._regex_dict:
if language_code == settings.LANGUAGE_CODE and not self.prefix_default_language:
regex_string = ''
else:
# start country changes
country_postfix = ''
if getattr(settings, 'SITE_COUNTRY', None):
country_postfix = '-{}'.format(settings.SITE_COUNTRY)
regex_string = '^%s%s/' % (language_code, country_postfix)
# end country changes
self._regex_dict[language_code] = re.compile(regex_string, re.UNICODE)
return self._regex_dict[language_code]
middleware.py - only very few lines changed, but had to replace the complete process_response.
from django.middleware.locale import LocaleMiddleware
from django.conf import settings
from django.conf.urls.i18n import is_language_prefix_patterns_used
from django.http import HttpResponseRedirect
from django.urls import get_script_prefix, is_valid_path
from django.utils import translation
from django.utils.cache import patch_vary_headers
class CountryLocaleMiddleware(LocaleMiddleware):
"""
This is a very simple middleware that parses a request
and decides what translation object to install in the current
thread context. This allows pages to be dynamically
translated to the language the user desires (if the language
is available, of course).
"""
response_redirect_class = HttpResponseRedirect
def process_response(self, request, response):
language = translation.get_language()
language_from_path = translation.get_language_from_path(request.path_info)
urlconf = getattr(request, 'urlconf', settings.ROOT_URLCONF)
i18n_patterns_used, prefixed_default_language = is_language_prefix_patterns_used(urlconf)
if (response.status_code == 404 and not language_from_path and
i18n_patterns_used and prefixed_default_language):
# Maybe the language code is missing in the URL? Try adding the
# language prefix and redirecting to that URL.
# start country changes
language_country = language
if getattr(settings, 'SITE_COUNTRY', None):
language_country = '{}-{}'.format(language, settings.SITE_COUNTRY)
language_path = '/%s%s' % (language_country, request.path_info)
# end country changes!
path_valid = is_valid_path(language_path, urlconf)
path_needs_slash = (
not path_valid and (
settings.APPEND_SLASH and not language_path.endswith('/') and
is_valid_path('%s/' % language_path, urlconf)
)
)
if path_valid or path_needs_slash:
script_prefix = get_script_prefix()
# Insert language after the script prefix and before the
# rest of the URL
language_url = request.get_full_path(force_append_slash=path_needs_slash).replace(
script_prefix,
'%s%s/' % (script_prefix, language_country),
1
)
return self.response_redirect_class(language_url)
if not (i18n_patterns_used and language_from_path):
patch_vary_headers(response, ('Accept-Language',))
if 'Content-Language' not in response:
response['Content-Language'] = language
return response
I'm using path converter in my django app like so:
# urls.py
from . import views
from django.urls import path
urlpatterns = [
path('articles/<str:collection>', views.ArticleView),
]
# views.py
#login_required
def ArticleView(request, collection):
print(collection)
if collection == "None":
articles_query = ArticleModel.objects.all()
...
This works fine in development for a url suck as : http://localhost:8000/articles/My Collection which gets encoded to http://localhost:8000/articles/My%20Collection, and is decoded properly in the ArticleView. However, in development, I have to edit the view like so to get it to work:
# views.py
import urllib.parse
#login_required
def ArticleView(request, collection):
collection = urllib.parse.unquote(collection)
print(collection)
if collection == "None":
articles_query = ArticleModel.objects.all()
...
Otherwise, the print(collection) shows My%20Collection and the whole logic in the rest of the view fails.
requirements.txt
asgiref==3.2.10
Django==3.1.1
django-crispy-forms==1.9.2
django-floppyforms==1.9.0
django-widget-tweaks==1.4.8
lxml==4.5.2
Pillow==7.2.0
python-pptx==0.6.18
pytz==2020.1
sqlparse==0.3.1
XlsxWriter==1.3.3
pymysql
What am I doing wrong here?
Thanks in advance!
The URL is being urlencoded which encodes spaces as %20. There are a number of other encodings. As you've discovered you need to decode that parameter in order to compare it to what you'd expect. As you've likely realized, if you have a value that actually wants The%20News and not The News, you have no recourse. To handle this people will create a slug field. Django has a model field for this in the framework.
This is typically a URL-friendly, unique value for the record.
Assuming you add a slug = models.SlugField() to ArticleModel, your urls and view can change into:
urlpatterns = [
# Define a path without a slug to identify the show all code path and avoid the None sentinel value.
path('articles', views.ArticleView, name='article-list'),
path('articles/<slug:slug>' views.ArticleView, name='article-slug-list'),
]
#login_required
def ArticleView(request, slug=None):
articles_query = ArticleModel.objects.all()
if slug:
articles_query = articles_query.filter(slug=slug)
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 am learning django by myself and created mini project for learning pupose and as example took a kidgarden.
I have a directorry sun and with two apps inside it accounts and equity apps.
In accounts i have templates with login signup htmls. This section works fine without any problem.
Your directory structure is still not clear. Do this. If it's works, please comment and I will explain what was wrong in your code.
in main dictionary (sun) urls.py, copy and paste this code.
from django.contrib import admin
from django.urls import path,include
from . import views
from accounts import views as account_views # added line
urlpatterns = [
path('admin/', admin.site.urls),
path('',views.HomePage.as_view(),name='home'),
path('accounts/',include('accounts.urls',namespace='accounts')),
path('accounts/',include('django.contrib.auth.urls')),
# path('kids/',views.TestPage.as_view(),name='kids'), # comment this url
path('kids/', account_views.home, name='kids'), # added url
path('thanks/',views.ThanksPage.as_view(),name='thanks'),
]
----- Edit ----
Initially, this was URL line. path('kids/',views.TestPage.as_view(),name='kids'). this was calling this function:
class ProductDetail(models.Model):
class TestPage(TemplateView):
template_name = 'equity/kids.html'
And your logic was written in this function (views in equity app):
def home(request):
stock = ChildForm()
if request.method == "POST":
stock = ChildForm(request.POST)
if stock.is_valid():
data = stock.save(commit=True)
name = data.name
context={
'name':name, }
else:
stock = ChildForm()
return render(request,'equity/kids.html',{'stock':stock})
return render(request,'equity/garden.html',context)
return render(request,'equity/kids.html',{'stock':stock)
You had to call right function which was in equity' views. So solution was this:
replace this path('kids/',views.TestPage.as_view(),name='kids')
with this:
path('kids/', account_views.home, name='kids')
importing views from equity is first thing.
from accounts import views as account_views
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