doesn't work when I change it this way
path('<str:gameslug>/player/<str:slug>',player_detail,name="player_detail"), //don't work 404 not found error
before changing (was working)
path('player/<str:slug>',player_detail,name="player_detail"), // was working
models.py
image=models.ImageField(null=True,blank=True,upload_to='player')
ERROR:Page not found (404)
i changed upload_to='player to
Now it is more clear
def create_path(instance, filename):
import os.path
return os.path.join(
instance.game.slug,
filename
)
now URL; media/gameslug/filepath
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 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've got an app deployed on Heroku using Django, and so far it seems to be working but I'm having a problem uploading new thumbnails. I have installed Pillow to allow me to resize images when they're uploaded and save the resized thumbnail, not the original image. However, every time I upload, I get the following error: "This backend doesn't support absolute paths." When I reload the page, the new image is there, but it is not resized. I am using Amazon AWS to store the images.
I'm suspecting it has something to do with my models.py. Here is my resize code:
class Projects(models.Model):
project_thumbnail = models.FileField(upload_to=get_upload_file_name, null=True, blank=True)
def __unicode__(self):
return self.project_name
def save(self):
if not self.id and not self.project_description:
return
super(Projects, self).save()
if self.project_thumbnail:
image = Image.open(self.project_thumbnail)
(width, height) = image.size
image.thumbnail((200,200), Image.ANTIALIAS)
image.save(self.project_thumbnail.path)
Is there something that I'm missing? Do I need to tell it something else?
Working with Heroku and AWS, you just need to change the method of FileField/ImageField 'path' to 'name'. So in your case it would be:
image.save(self.project_thumbnail.name)
instead of
image.save(self.project_thumbnail.path)
I found the answer with the help of others googling as well, since my searches didn't pull the answers I wanted. It was a problem with Pillow and how it uses absolute paths to save, so I switched to using the storages module as a temp save space and it's working now. Here's the code:
from django.core.files.storage import default_storage as storage
...
def save(self):
if not self.id and not self.project_description:
return
super(Projects, self).save()
if self.project_thumbnail:
size = 200, 200
image = Image.open(self.project_thumbnail)
image.thumbnail(size, Image.ANTIALIAS)
fh = storage.open(self.project_thumbnail.name, "w")
format = 'png' # You need to set the correct image format here
image.save(fh, format)
fh.close()
NotImplementedError: This backend doesn't support absolute paths - can be fixed by replacing file.path with file.name
How it looks in the the console
c = ContactImport.objects.last()
>>> c.json_file
<FieldFile: protected/json_files/data_SbLN1MpVGetUiN_uodPnd9yE2prgeTVTYKZ.json>
>>> c.json_file.name
'protected/json_files/data_SbLN1MpVGetUiN_uodPnd9yE2prgeTVTYKZ.json'
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
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 :)