5
我有一個Zend表格來添加一些東西到數據庫。然後我想用這個表單來編輯我添加到數據庫中的內容。是否有任何可能使用這種形式(從數據庫填充它,並顯示它???) 我在我的控制器有這樣的:ZEND,編輯表格
public function editAction() {
if (Zend_Auth::getInstance()->hasIdentity()) {
try {
$form = new Application_Form_NewStory();
$request = $this->getRequest();
$story = new Application_Model_DbTable_Story();
$result = $story->find($request->getParam('id'));
// $values = array(
// 'title' => $result->title,
// 'story' => $result->story,
//);
if ($this->getRequest()->isPost()) {
if ($form->isValid($request->getPost())) {
$data = array(
'title' => $form->getValue("title"),
'story' => $form->getValue("story"),
);
$where = array(
'id' => $request->getParam('id'),
);
$story->update($data, $where);
}
}
$this->view->form = $form;
$this->view->titleS= $result->title;
$this->view->storyS= $result->story;
} catch (Exception $e) {
echo $e;
}
} else {
$this->_helper->redirector->goToRoute(array(
'controller' => 'auth',
'action' => 'index'
));
}
}
筆者認爲:
<?php
try
{
$tmp = $this->form->setAction($this->url());
//$tmp->titleS=$this->title;
//$tmp->storyS=$this->story;
//echo $tmp->title = "aaaaa";
}
catch(Exception $e)
{
echo $e;
}
當我嘗試改變這個視圖中的東西我的意思是給任何不同的值然後NULL我有錯誤,我不能這樣做,因此是否有任何可能重用這種形式?或不?
謝謝!
是的它的工作原理!非常感謝你! – canimbenim 2011-01-29 23:13:05