This worked perfectly, thanks!
How would I serve up the image if I was broadcasting an XML service
API?
I want my service to respond to a request with images that can be
displayed on their website.
Thanks!
On Jun 15, 1:57 am, mdipierro <mdipie...@cs.depaul.edu> wrote:
> I marked with ## the lines that I added, changed. Hope it is self-
> explanatory.
> Let me know if this fixes it.
>
> def add():
> form = SQLFORM(db.apps)
> if form.accepts(request.vars, session):
> redirect(URL(r=request, c='default', f='index'))
> records=db(db.apps.id>0).select() ####
> return dict(form = form,records=records) ###
>
> def download(): ###
> return response.download(request,db) ###
>
> And in the view
>
> {{for app in records:}}
> <div>
> <img src="{{=URL(r=request,
> f='download', args=app.screenshot)}}"> ####
> {{=A(app.name, _href=URL(r=request,
> f=app.link))}}
> <span>by {{=app.author}}</span>
> <br/>
> <span>{{=app.description}}</span>
> </div>
> {{pass}}
>
> On Jun 14, 10:28 pm, Michael <michael.no...@gmail.com> wrote:
>
>
>
> > I have a model that has:
>
> > db.Field('screenshot', 'upload'))
>
> > In my controller i have:
>
> > def add():
>
> > form = SQLFORM(db.apps)
>
> > if form.accepts(request.vars, session):
> > redirect(URL(r=request, c='default', f='index'))
>
> > return dict(form = form)
>
> > and in my view I have:
>
> > {{for app in records:}}
> > <div>
> > <img src="{{=URL(r=request, c='uploads',
> > f=app.screenshot)}}">
> > {{=A(app.name, _href=URL(r=request,
> > f=app.link))}}
> > <span>by {{=app.author}}</span>
> > <br/>
> > <span>{{=app.description}}</span>
> > </div>
> > {{pass}}
>
> > this returns a broken image. anyone know how to display an image from
> > the upload of a form?
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