Don't. This would break lots of stuff.
Instead create your own custom download function
def donwload2():
row_id=request.args(0).split('.')[0]
image = db.yourtable[row_id].image
request.args=[image]
return response.download(request,response,db)
On Jan 6, 12:40 am, Mikko <[email protected]> wrote:
> Hello,
>
> I have db table: SQLFIELD( row_id, string), SQLFIELD(image, upload)
>
> how do I make image name look like row_id + ".jpeg"?
>
> I'm now using SQLFORM.
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