I want my controller to return a UPC barcode as action with one
argument, like so: mycontroller/barcode/012345678901
It works fine if I read a file object, like so:
def barcode(self, data):
f = open('barcode.jpg', 'r')
barcode_('upc-a', str(data)).save(f, format='JPEG')
response.headers['Content-Type'] = 'image/jpg'
f.close()
f.open('barcode.jpg')
return f.read()
But I don't want to do that because then I have a bunch of temp files
to clean up. But using StringIO just gives me the URL in the browser
window (no image):
def barcode(self, data):
import StringIO
f = StringIO.StringIO()
barcode_('upc-a', str(data)).save(f, format='JPEG')
response.headers['Content-Type'] = 'image/jpg'
return f.buff
Any tips on how to render an image like this without writing/reading
the persisted file would be helpful. Thanks.
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