cakePHP 1.3：使用Ajax呈現問題

我想在我的ajax響應中使用一個元素作爲字符串。cakePHP 1.3：使用Ajax呈現問題

，所以我已經把下面的代碼在我的控制器行動

$view = new View($this, false); 
$view->viewPath = 'elements/private_challenge/';    
$res = $view->render('supporters_list_supporter');

現在我supporters_list_supporter.ctp我已經把如下一種形式：

echo $this->Form->create('RequestFrHelpReply',array('action'=>'add_helpreq_reply','class'=>'form','id' => 'helprequestreplyaddfrm','name'=>'helprequestreplyaddfrm')); 

echo $this->Form->hidden('PrivateChallengeRequestHelp.ReqID',array('value'=>$supporterlist['RequestFrHelp']['REQID'])); 
echo "<div class='alignleft'>"; 
echo $this->Form->input('PrivateChallengeRequestHelp.txt_helpreq_reply', array('label'=>false,'type' => 'textarea', 'id'=>'txt_helpreq_reply','escape' => false, 'div' => false, "class"=>"textarea request-help-textarea",'onKeyDown'=>'textCounterComment(this.form.txt_helpreq_reply,this.form.remLenth,'.Configure::read('CommentSetting.comment_body_max_limit').',\'remLenthDiv\');', 'onKeyUp'=>'textCounterComment(this.form.txt_helpreq_reply,this.form.remLenth,'.Configure::read('CommentSetting.comment_body_max_limit').',\'remLenthDiv\');')); 
    echo "</div>"; 
    echo '<span class="request-arrow"></span><span class="user-name"> : '.$html->get_users_username($session->read('Auth.User.id')).'</span>'; 
echo $html->link(__l('Reply'),'#',array('id'=>'helpreq_reply_link','class'=>'request-help-button alignleft','title'=>__l('Reply'),'escape'=>false,'onclick'=>'helpreq_reply(\'reply\');return false;'));    
echo "</div>";  
echo $this->Form->end();

因此，它給我的錯誤有關形式助手失蹤，所以任何人都可以幫助我如何添加形式助手這個exetrnal對象？

更新：錯誤消息：

$view = stdClass 
    stdClass::$modelScope = false 
    stdClass::$entityPath = "RequestFrHelpReply."</pre><pre class="stack-trace">Helper::setEntity() - CORE\cake\libs\view\helper.php, line 422 
    FormHelper::create() - CORE\cake\libs\view\helpers\form.php, line 217 
    include - APP\views\elements\private_challenge\help_request_form.ctp, line 21

來源

2012-03-21 chetanspeed511987