yes
something like that ;)
thanks Massimo
On Friday, July 17, 2015 at 3:13:56 AM UTC-5, Massimo Di Pierro wrote:
>
> Something like this?
>
> db.define_table('item', Field('name'),Field('value','double'))
>
> def index():
> c = lambda id: "c%s" % id
> table = TABLE()
> item = db(db.item).select():
> for item in items:
> table.append(TR(LABEL(item.name),INPUT(_name=c(item.id
> ),value=item.value)))
> form = FORM(table)
> if form.process().accepted:
> for item in items:
> newvalue = form.vars[c[item.id]]
> if newvalue != str(item.value):
> item.update_record(value=newvalue)
> response.flash = 'saved'
> return dict(form=form)
>
>
>
> On Thursday, 16 July 2015 17:30:23 UTC-5, JorgeH wrote:
>>
>> Hello
>>
>> I need to generate a grid like
>>
>>
>> <https://lh3.googleusercontent.com/-Y_uOwCxGkY0/Vaguw-0K1DI/AAAAAAAAAJM/4dkpl2IQORc/s1600/Captura.PNG>
>>
>>
>>
>>
>> table definition is:
>>
>> db.define_table ('produccion',
>> Field ('proyecto', 'reference proyectos'),
>> Field ('month', 'date'),
>> Field ('ton_presup', label='Budgeted Tons'),
>> Field ('ton_adic', label='Additional Tons'),
>>
>> )
>>
>>
>> I have currently generated the form in the view only, iterating rows, but
>> want to do it from the controller using form factory for validation
>> purposes.
>>
>>
>>
>> the number of rows and starting month may vary.
>>
>> How can I use a list item as a variable name, and by the way is this the
>> best approach?
>>
>>
>>
>>
>> Any hints?
>>
>> thanks
>>
>
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