php - 我们如何在 drupal 7 中使用 ajax 回调函数将下拉值作为参数传递？

Question

我想将 ajax 回调函数中的下拉值作为参数 ($user_id,$selected) onchange 传递，但我没有在这个 ajax 回调函数 onchange 事件中获得 $selected 值**

Answer 1

$form['changethis'] = array( 
'#type' => 'select', 
'#title' => t('Rece/Paid'), 
'#description' => t('See Records'), 
'#options' => $options, 
'#default_value' => $selected, 
'#ajax' => array( 
  'callback' => 'ajax_callback', 
  'wrapper' => 'replace_textfield_div', 
  'event' => 'change', 
  'progress' => array( 'type' => 'none', ),
 ), 
);

以上添加是为了便于阅读。

您的选择是否发布了某些内容但没有得到结果？或者它根本没有发布任何东西？

您的代码看起来不错，但您确定要使用 ajax 回调“ajax_callback”吗？这不应该是您自己的返回部分表单的函数吗？

另外，我不确定'progress' => array( 'type' => 'none', ),文档说只有“throbber”和“bar”是有效的。

尝试这样的事情

function my_module_create_form($form, $form_state){

  // check form_state here to see if the form has been submitted
  // via ajax and change if needed.

  $options = array(0, 1, 2);
  $selected = 0;

  $form = array();
  $form['changethis'] = array( 
  '#type' => 'select', 
  '#title' => t('Rece/Paid'), 
  '#description' => t('See Records'), 
  '#options' => $options, 
  '#default_value' => $selected, 
  '#ajax' => array( 
    'callback' => 'my_module_ajax_callback', 
    'wrapper' => 'replace_textfield_div', 
    'event' => 'change', 
   ), 
  );
  return $form;
}

function my_module_ajax_callback($form, $form_state) {

  // because this is called from change event on select, 
  // it wont rebuild the form (call my_module_create_form).
  // If you want it to you have to set $form_state['rebuild']
  // (or it may be $form['rebuild'], I can't remember which one.)      

  // return form part
  return $form['changethis'];
}

Answer 2

使用 jQuery：$("#yourdropdownid option:selected").text();