Can't you just cache the result of the f.read() call, delete the file
and then return from the call?
f.close()
f.open('barcode.jpg')
tmp = f.read()
os.remove('barcode.jpg') # should really create a unique temp
name to avoid concurrency problems
On Dec 2, 2:33 pm, "[email protected]"
<[email protected]> wrote:
> 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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