Question

我在页面中显示元素时出现问题。根据角色的选择 - 我需要显示|隐藏选择元素。所以我喜欢这样的元素：

$form['field_test_field']['#access'] = 0;

但是当状态发生变化时我无法通过绑定来恢复它。那就是我的片段：

function seven_form_user_register_form_alter(&$form, &$form_state, $form_id) {
    $form['account']['roles']['#ajax'] = array(
        'callback' => 'seven_dynamic_form_ajax_callback',
        'wrapper' => 'replace_droplist',
        'event' => 'change',
    );

    $form['field_test_field']['#prefix'] = '<div id="replace_droplist">';
    $form['field_test_field']['#suffix'] = '</div>';
    $form['account']['roles']['#type'] = 'radios';

    $checkedRole = isset($form_state['values']['roles'])
        ? $form_state['values']['roles'] : 0;

    if ($checkedRole == 0) {
        $form['field_test_field']['#access'] = 1;
    } elseif ($checkedRole == 4) {
        $form['field_test_field']['#access'] = 0;
    } elseif ($checkedRole == 3) {
        $form['field_test_field']['#access'] = 1;
    }
};

function seven_dynamic_form_ajax_callback($form, $form_state) {
    return $form['field_test_field'];
}

感谢。

Answer 1

<code>
$form['enter_amt'] = array(
    '#type' => 'checkbox',
    '#title' => t('Enter Billable Amount'),
  );
  $form['billable_ctc'] = array(
    '#type' => 'textfield',
    '#title' => t('Billable CTC'),
    '#maxlength' => '15',
    '#field_prefix' => t('Rs.'),
    '#states' => array('invisible' => array(':input[name="enter_amt"]' =>         array('checked' => TRUE), ), ),
  );
  $form['billing_rate'] = array(
    '#type' => 'textfield',
    '#title' => t('Billing rate'),
    '#maxlength' => '7',
    '#field_suffix' => '%',
    '#size' => 3,
    '#attributes' => array('class' => array('auto-width')),
    '#states' => array('invisible' => array(':input[name="enter_amt"]' => array('checked' => TRUE), ), ),
  );
</code>

此代码可让我根据复选框隐藏和显示特定的表单元素。

显示/隐藏元素ajax Drupal

1 个答案: