Django从数据库提供图像

Question

I am adding trying to get an app to serve image fields in my database to display in my template. I was wondering if there was a way to bypass having to use a global media dir/url setting, for the sake of organization. In my modal i have set and upload to option. Id like to use a url mapping for the app with regex named groups to tell my view what the path and the image name is. concatenate that into a string and serve back the file. If this is a terrible idea, or is just grossly inefficient, I'll just do something else. Any help is greatly appreciated. Here is what I have so far.

Template:`

{% for category in items %}
                <div class="col-sm-4 col-lg-4 col-md-4">
                    <div class="thumbnail">
                        <img src="{{ category.picture }}" />
                        <div class="caption">
                            <h4 class="pull-right">${{ category.price }}</h4>

                            <h4><a href="#">{{ category.itemName }}</a>

                            </h4>
                        </div>

                    </div>
                </div>
{% endfor %}

Models.py:

from django.db import models

class StoreCategory(models.Model):
    categoryName = models.CharField(max_length=128)
    def __unicode__(self):
            return self.categoryName

class StoreItem(models.Model):

    category = models.ForeignKey(StoreCategory)
    itemName = models.CharField(max_length=128)
    description = models.CharField(max_length=1024)
    price = models.DecimalField(max_digits=16,decimal_places=2)
    quantity = models.IntegerField(default=0)
    picture = models.ImageField(upload_to='store_images', blank=False)
    def __unicode__(self):
            return  self.itemName

Urls.py:

urlpatterns = patterns('',
url(r'^$', views.webstore, name='webstore'),
url(r'^FiberArts/$', views.FiberArts, name='FiberArts'),
url(r'^FiberArts/(?P<directory>[\w]+)/(?P<image_name>[\w]+)$',views.getImage)
)

View: ( currently this is never called, problem with regex?)

def getImage(request, directory, image_name):
    imagelocation = directory + '/' + image_name
    print imagelocation
    image_data = open(imagelocation, "rb").read()

    return HttpResponse(image_data, mimetype="image/png")

Edit: Here is the error I'm getting:

[06/Apr/2014 15:48:35] "GET /webstore/FiberArts/store_images/product_150_1.jpg HTTP/1.1" 404 3743

Answer 1

Change that url regex to this (?P<directory>[\\w]+)/(?P<image_name>[\\w]+\\.[\\w]+)$' and fix your view to accept the right number of arguments. You should now receive the right image_name inside your view.