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Home/ Questions/Q 7787619
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T20:43:45+00:00 2026-06-01T20:43:45+00:00

I’m converting my site over to using the blobstore for image serving and am

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I’m converting my site over to using the blobstore for image serving and am having a problem. I have a page with a large number of images being rendered dynamically (through jinja), and the only data available are entity keys that point to image objects that contain the relevant serving url.

Previously each image had a url along the lines of "/show-image?key={{image_key}}", which points to a request handler along the lines of this:

def get(self):
  imageInfo = db.get(self.request.args.get("key"))
  imagedata = imageInfo.data // the image is stored as a blob in the normal datastore
  response = Response()
  response.data = imagedata
  response.headers['Content-Type'] = imageInfo.type
  return response

My question is: How can I modify this so that, rather than returning a response with imageInfo.data, I return a response with imageInfo.saved_serving_url (generated from get_serving_url when the image object was created). More importantly, is this even a good idea? It seems like converting the saved_serving_url back into data (eg using urllib.fetch) might just counteract the speed and efficiency of using the high-speed datastore in the first place?

Maybe I should just rewrite my code so that the jinja template has direct access to the serving urls of each image. But ideally I’d like to avoid that due to the amount of parallel lists I’d have to pass about.

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  1. Editorial Team
    Editorial Team
    2026-06-01T20:43:47+00:00Added an answer on June 1, 2026 at 8:43 pm

    why not returning the serving url instead of the imagedata?

    <img src="/show-image?key={{image_key}}" />
    
    def get(self):
      imageInfo = db.get(self.request.args.get("key"))
      return imageInfo.saved_serving_url
    
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