Hi there,
I'm trying to create a photo gallery by storing photos in a gallery
database and pulling them out when appropriate.
However, I ran into the problem of how to publicly expose the pictures in
urls. In my template, I return a the url of the photo which turns out to be
images/photo.jpg. However in my development server, when I put in the url
192.0.0.1:8000/images/photo.jpg, I can't view the image. Is there any way
to make the url/image accessible through the url provided by the object
without manually tampering with urls.py? If not, why does Django provide
the .url field? Is this the only a recommended url field then?
My View:
def gallery(request):
base_gallery = gall.objects.get(name="base")
base_list = base_gallery.photos.all()
return render_to_response('gallery.html',
{'list_of_images':base_list},
context_instance = RequestContext(request),
)
Template:
{% extends "index.html" %}{% block body_content %}
{% for photo in list_of_images %}
{{ photo.image.url }}{% endfor %}{% endblock %}
Model:
from django.db import models
class photo(models.Model):
name = models.CharField(max_length=50,blank=True,null=True)
source_descr = models.CharField(max_length=100,blank=True,null=True)
image = models.ImageField(upload_to='images')
class gallery(models.Model):
name = models.CharField(max_length=50)
photos = models.ManyToManyField(photo)
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