我正在使用 PHP 5.5 在本地计算机上运行 Drupal 6,当我尝试添加通过模块创建的内容类型的节点时,我得到了不受支持的操作数类型。
该错误正在被触发function drupal_render
,可以在 中找到commons.inc
。线是$elements += array('#title' => NULL, '#description' => NULL);
。我做了一个var_dump
并发现由于某种原因我的元素是一个布尔值,而不是一个数组。我不知道为什么。
这是我创建表单的模块文件
<?php
// $ID$
function jokes_node_info(){
return array(
'jokes' => array(
'name' => t('jokes'),
'module' => 'jokes',
'description' => t('Tell use your joke'),
'has_title' => true,
'title_label' => t('Title'),
'has_body', true,
'body_label' => t('jokes'),
'min_word_count' => 2,
'locked' => true
)
);
}
//only admin can create jokes
function jokes_menu_alter(&$callback){
if(!user_access('administer nodes')){
$callback['node/add/jokes']['access callback'] = false;
unset($callback['node/add/jokes']['access arguments']);
}
}
//create permissions
function jokes_perm(){
return array(
'create jokes',
'edit own jokes',
'edit any jokes',
'delete own jokes',
'delete any jokes'
);
}
//control access
function jokes_access($op, $node, $account){
$is_author = $account->uid == $node->uid;
switch ($op) {
case 'create':
return user_access('create jokes', $account);
case 'edit':
return user_access('edit own jokes', $account) && $is_author || user_access('edit any jokes', $account);
case 'delete':
return user_access('delete own jokes', $account) && $is_author || user_access('delete any jokes', $account);
default:
break;
}
}
function jokes_form($node){
$type = node_get_types('type', $node);
$form['title'] = array(
'#type' => 'textfield',
'#title' => check_plain($type->title_label),
'#required' => true,
'#default_value' => $node->title,
'#weight' => -5,
'#maxlength' => 255
);
$form['body_filter']['body'] = array(
'#type' => 'textarea',
'#title' => check_plain($type->body_label),
'#default_value' => $node->body,
'#rows' => 7,
'required' => true
);
$form['body_filter']['filter'] = filter_form($node->format);
$form['punchline'] = array(
'#type' => 'textfield',
'#title' => t('Punchline'),
'#required' => true,
'#default_value' => isset($node->punchline) ? $node->punchline : '',
'#weight' => 5
);
return $form;
}
?>