It probably would be more beneficial to look at the validators.py class IS_IN_SET(Validator):
add print statements to see why it isn't working. On Thu, Mar 1, 2012 at 7:05 AM, Cliff <[email protected]> wrote: > Running 1.99.4 > > I have not been able to get the IS_IN_SET validator to work properly. > I don't know why this should be. > > When used with SQLFORM to edit a record, the select field refuses to > show the current value of the field. Instead it shows the first value > in the list or the zero value for the field. > > So I made an alternative widget. The code consists of two functions, > below. > > If you are having similar problems, feel free to use it. > > The _style definition at the end of the call is there to make sure I > was seeing my widget instead of the default. It is not necessary for > the widget to work. > > ############################################################ > > def is_it_selected(current_value, option, multiple): > if len(current_value) == 1 or not multiple: > current_value=current_value[0] > if current_value==option: > out = 'selected' > else: > out = None > else: > if option in current_value: > out='selected' > else: > out=None > return out > > def my_select_widget(field,value): > # diagnostics uncomment to learn > #print 'value' > #print value > #print 'type(value)' > #print type(value) > > f = str(field) > f_name = f.split('.')[-1] > f_id = f.replace('.', '_') > f_requires = db[request.controller][f_name].requires > the_set = f_requires.theset > the_multiple = f_requires.multiple > option_set = [] > for t in the_set: > is_selected = is_it_selected(value, t, the_multiple) > option_set.append(OPTION(t, _selected = is_selected)) > return SELECT(option_set, _name='status', _id='buckslips_status', > _style='color:red;' > ) -- -- Regards, Bruce Wade http://ca.linkedin.com/in/brucelwade http://www.wadecybertech.com http://www.warplydesigned.com http://www.fitnessfriendsfinder.com
