dynamically fill list?

Hey,

Sorry for the delay – I didn’t get a notification.

You’ll need to modify the form object using the gform_pre_render filter (and the other similar filters).

Example below shows how to do this — for FORM ID 1 FIELD ID 20 and COLUMN LABEL ‘Column 1’. Note how the values are a comma separated string (and we can’t define the option value separately at the moment).

(note I’ve assumed you’re working with a multi-column list field)

add_filter( 'gform_pre_render', 'my_gform_pre_render', 10, 1 );
add_filter( 'gform_pre_validation', 'my_gform_pre_render', 10, 1 );
add_filter( 'gform_admin_pre_render', 'my_gform_pre_render', 10, 1 );
add_filter( 'gform_pre_submission_filter', 'my_gform_pre_render', 10, 1 );

function my_gform_pre_render( $form ) {
	if ( GFCommon::is_form_editor() || 1 != $form['id'] ) {
		return $form;
	}

	if ( is_array( $form ) || is_object( $form ) ) {

		foreach ( $form['fields'] as &$field ) {  // for all form fields
			$field_id = $field['id'];

			if ( 20 != $field_id ) {
				break;
			}

			if ( 'list' == $field->get_input_type() ) {

				$has_columns = is_array( $field->choices );

				if ( $has_columns ) {
					foreach ( $field->choices as $key => &$choice ) { // for each column
						$isDropDown = rgar( $choice, 'isDropDown' );
						$column = rgars( $field->choices, "{$key}/text" );
						if ( $isDropDown && 'Column 1' == $column ) {
							$choices = 'Option 1, Option 2, Option 3';
							$choice['isDropDownChoices'] = $choices;
						}
					}
				}
			}
		}
	}
	return $form;
}

• This reply was modified 7 years, 10 months ago by ovann86.