well, i've done, but it seems more like a hack:
def show():
response.headers['Content-Type']='image/jpeg'
img = db(db.images.id == request.args[0]).select()[0]
import os
return response.stream(open(os.getcwd() + '/applications/' +
request.application + '/uploads/' + img.image, 'rb'))
this allows me to get my jpg images through
http://mysite/myapp/images/show/4
On 31 Gen, 17:44, pistacchio <[email protected]> wrote:
> so, if if i got it correctly, wouldn'thttp://myhost/myapp/mycontrol/download/4
> return my image with id 4? this is what i want, but i get a 404
> instead..
>
> On 31 Gen, 17:33, mdipierro <[email protected]> wrote:
>
>
>
> > This code
>
> > {{=URL(r=request, f='download', args= IMAGE)}}
>
> > simply generates
>
> > /<app>/default/download/<filename>
>
> > prepend http://<hostname>
>
> > and you get a static URL.
>
> > Within the app itself you should use the {{=URL(..)}} function because
> > the app may be renamed and because it may be behind routes. When your
> > app is published online you can use the static link to reference the
> > download action from other apps.
>
> > On Jan 31, 10:27 am, pistacchio <[email protected]> wrote:
>
> > > hi!
> > > i've seen that i can retrieve uploaded images with
>
> > > {{=URL(r=request, f='download', args= IMAGE)}}
>
> > > coupled with the standard "download" method.
>
> > > now, if i want to statically save a reference to the image (like
> > > saving the string "<img src=/myapp/mycontroller/download/
> > > mycontroller.imagexxxxxxxxxxxxxxxx" etc, how to do that?
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