Can you provide an example of IS_IN_SET not working so we can fix it? By
the way, there was a change in IS_IN_SET in trunk and that may have fixed
your problem.
On Thursday, 1 March 2012 09:05:51 UTC-6, Cliff 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;'
> )
