Thank you, that was the solution I've seached for.
On 2 Jun, 13:21, "hamdy.a.farag" wrote:
> assuming you've
>
> def index():
> form = SQLFORM(db.tab)
> if form.accepts(request.vars, session):
> response.flash = 'x'
> records = db().select(db.tab.ALL)
> return dict(form=fo
oh btw
the image part should look like:
{{=IMG(_src=URL(r=request, c='default', f='download',
args=[record.x]))}}
On Jun 2, 2:21 pm, "hamdy.a.farag" wrote:
> assuming you've
>
> def index():
> form = SQLFORM(db.tab)
> if form.accepts(request.vars, session):
> response.flash = 'x
assuming you've
def index():
form = SQLFORM(db.tab)
if form.accepts(request.vars, session):
response.flash = 'x'
records = db().select(db.tab.ALL)
return dict(form=form, records=records)
then in view you can do something like:
{{extend 'layout.html'}}
{{=form}}
Not really. The information about the filename must be coded into the
filename in the uploadfolder. But how to get the filename?
On 2 Jun, 11:10, Thadeus Burgess wrote:
> Taken fromhttp://web2py.com/book/default/section/7/2?search=original+filename
>
> Occasionally you may want to store the origi
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