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copied!<p>You should have an <code><img></code> tag in your template that has a <code>src</code> attribute that contains a URL that is served by your application and delivers the data for the image. For example, let's say that you store images in a model called Image:</p> <pre><code>class Image(db.Model): filename = db.StringProperty() # The name of the uploaded image mime_type = db.StringProperty() # The mime type. content = db.BlobProperty() # The bytes of the image def load(id): # Load an entity from the database using the id. Left as an # exercise... def link_id_for(self): "Returns an id that uniquely identifies the image" return self.key().id() </code></pre> <p>In the controller/request handler code that renders the page that contains the image, you would pass the id that is returned by <code>link_id_for</code> to the template, and the template would have your image tag, something like this:</p> <pre><code><img src="/images/show/{{image_id}}"> </code></pre> <p>You would have a request handler that handles requests to <code>/images/show/id</code>. You would use <em>id</em> to get the Image entity out of the datastore and the send it back in the response, something like this:</p> <pre><code>found_image = Image.load(id) response.headers['Content-Type'] = str(found_image.mime_type) response.out.write(found_image.content) </code></pre> <p>Obviously, you'd have to adapt the particulars of the code to your current application structure and conventions, but that's the heart of it: use an <code>img</code> tag with a <code>src</code> that points at your application; your application includes a request handler the delivers the bytes and a <code>Content-Type</code> header.</p>

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