问题
I am trying to find the most efficient way of displaying an image using django's template context loader. I have a static dir within my app which contains the image 'victoryDance.gif' and an empty static root dir at the project level (with settings.py
). assuming the paths within my urls.py
and settings.py
files are correct. what is the best view?
from django.shortcuts import HttpResponse
from django.conf import settings
from django.template import RequestContext, Template, Context
def image1(request): # good because only the required context is rendered
html = Template('<img src="{{ STATIC_URL }}victoryDance.gif" alt="Hi!" />')
ctx = { 'STATIC_URL':settings.STATIC_URL}
return HttpResponse(html.render(Context(ctx)))
def image2(request): # good because you don't have to explicitly define STATIC_URL
html = Template('<img src="{{ STATIC_URL }}victoryDance.gif" alt="Hi!" />')
return HttpResponse(html.render(RequestContext(request)))
def image3(request): # This allows you to load STATIC_URL selectively from the template end
html = Template('{% load static %}<img src="{% static "victoryDance.gif" %}" />')
return HttpResponse(html.render(Context(request)))
def image4(request): # same pros as image3
html = Template('{% load static %} <img src="{% get_static_prefix %}victoryDance.gif" %}" />')
return HttpResponse(html.render(Context(request)))
def image5(request):
html = Template('{% load static %} {% get_static_prefix as STATIC_PREFIX %} <img src="{{ STATIC_PREFIX }}victoryDance.gif" alt="Hi!" />')
return HttpResponse(html.render(Context(request)))
thanks for answers These views all work!
回答1:
If you need to render an image read a bit here http://www.djangobook.com/en/1.0/chapter11/ and use your version of the following code:
For django version <= 1.5:
from django.http import HttpResponse
def my_image(request):
image_data = open("/path/to/my/image.png", "rb").read()
return HttpResponse(image_data, mimetype="image/png")
For django 1.5+ mimetype
was replaced by content_type
(so happy I'm not working with django anymore):
from django.http import HttpResponse
def my_image(request):
image_data = open("/path/to/my/image.png", "rb").read()
return HttpResponse(image_data, content_type="image/png")
Also there's a better way of doing things!
Else, if you need a efficient template engine use Jinja2
Else, if you are using Django's templating system, from my knowledge you don't need to define STATIC_URL as it is served to your templates by the "static" context preprocessor:
TEMPLATE_CONTEXT_PROCESSORS = (
'django.contrib.auth.context_processors.auth',
'django.core.context_processors.debug',
'django.core.context_processors.i18n',
'django.core.context_processors.static',
'django.core.context_processors.media',
'django.core.context_processors.request',
'django.contrib.messages.context_processors.messages',
)
回答2:
In your last example (image5) you should use {{ STATIC_PREFIX }}
instead of {% STATIC_PREFIX %}
STATIC_PREFIX
is variable, not a tag
回答3:
To avoid defining STATIC_URL
explicitly, you can use a RequestContext when rendering your template. Just make sure django.core.context_processors.static
is in your TEMPLATE_CONTEXT_PROCESSORS
setting.
from django.template import RequestContext
...
return HttpResponse(html.render(RequestContext(request, ctx)))
Alternatively, you could use the static template tag.
html = Template('<img src="{% static "victoryDance.gif" %} alt="Hi!" />')
A third option is the get_static_prefix template tag.
来源:https://stackoverflow.com/questions/11577681/a-simple-view-to-display-render-a-static-image-in-django