Thanks, Ian - that worked!

@Marco:  Thanks for the alternative technique - I think I like the
idea of using StringIO and not creating/ deleting the file.

On Dec 2, 4:20 pm, Ian Wilson <[email protected]> wrote:
> Seems that you want something like this when using StringIO (in python 2.6
> at least):
>
> try:
>     return f.getvalue()
> finally:
>     f.close()
>
> On Thu, Dec 2, 2010 at 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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