You can change the field requires in the controller depending on whether
the field should be shown doing something like this
def index():
db.purchase.coupon_code.show_if = (db.purchase.have_coupon==True)
if request.vars.have_coupon:
db.purchase.coupon_code.requires = IS_NOT_EMPTY()
else:
db.purchase.coupon_code.requires = IS_EMPTY()
form = SQLFORM(db.purchase).process()
return dict(form = form)
db.purchase.coupon_code.show_if = (db.purchase.have_coupon==True)
A quinta-feira, 3 de dezembro de 2020 à(s) 07:05:55 UTC,
[email protected] escreveu:
> All,
>
> I am trying to get a solution to the above problem, I am using a
> onvalidation function. If have_coupon is checked then I call IS_NOT_EMPTY
> and set the form.errors.coupon_code manually. This is not ideal but it
> works.
>
> However I cannot seem to get a work around for a dropdown input i.e.
> IS_IN_SET as this is needed on the model to create the correct input but
> inturn fails the processing of the form as onvalidation doesnot get called.
>
> Any help would be great.
>
> Regards,
> James
>
>
> On Wednesday, December 2, 2020 at 12:35:15 PM UTC+10 James O' Driscoll
> wrote:
>
>> Hello all,
>>
>> I have a question regarding conditional fields and using validators.
>>
>> Take the example from the book with a slight change:
>>
>> db.define_table('purchase', Field('have_coupon', 'boolean'),
>> Field('coupon_code', requires=IS_NOT_EMPTY()))
>>
>> with controller:
>>
>> def index():
>> db.purchase.coupon_code.show_if = (db.purchase.have_coupon==True)
>> form = SQLFORM(db.purchase).process()
>> return dict(form = form)
>>
>> show_if works well but if i need validators in place for when say
>> have_coupon is true, to validate coupon_code I get an error when
>> have_coupon is false as coupon_code is empty.
>>
>> I would prefer to keep the validator on the model, is there a way to
>> accomplish this.
>>
>> Regards,
>> James
>>
>>
--
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