I guess, you will require to create a __toString or __string function
which will return the name instead of id's for choice menus.
Additionally check if you tables have some data to show in choice
menus. the table which is available as choice data to other table
should have some data or ___string
how did you solve, share... so that everybody can get to know
On Aug 31, 2:55 pm, asim nizam asim...@gmail.com wrote:
Thanks i have solved problem!!
On Mon, Aug 31, 2009 at 3:47 PM, krishan milepe...@gmail.com wrote:
I guess, you will require to create a __toString or __string function
yes sir it was to __toStrig method and you discusssed below clearly thanks.
On Mon, Aug 31, 2009 at 5:07 PM, krishan milepe...@gmail.com wrote:
how did you solve, share... so that everybody can get to know
On Aug 31, 2:55 pm, asim nizam asim...@gmail.com wrote:
Thanks i have solved
Problem might be somewhere in \lib\form\ and model's form base class file.
check if $this-setWidgets() function is provided with appropriate choice
widget. May be this can help..
On Wed, Aug 26, 2009 at 2:15 PM, sunny asim...@gmail.com wrote:
I get the following error message in the form!
What's your form code like?
On Wed, Aug 26, 2009 at 8:59 PM, Krishan .Gmilepe...@gmail.com wrote:
Problem might be somewhere in \lib\form\ and model's form base class file.
check if $this-setWidgets() function is provided with appropriate choice
widget. May be this can help..
On Wed, Aug
here is my form code
class HrTblJobApplicationForm extends BaseHrTblJobApplicationForm
{
public function configure()
{
$this-setWidgets(array(
'application_id'= new sfWidgetFormInputHidden(),
'vacancy_id'= new
Just guessing: are you dealing with complex primary keys on any propel-
choice-related table? I think you might be getting an array as an
index, when a simple string or number is actually expected.
--
José Nahuel Cuesta Luengo
El 26/08/2009, a las 06:15, asim nizam asim...@gmail.com escribió:
