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zend-framework - 通过 Zend Paginator 分页给出错误

转载 作者:行者123 更新时间:2023-12-04 06:44:03 26 4
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我在我的应用程序中使用 zend paginator 进行分页,这是这样做的操作

public function listAction(){
$registry = Zend_Registry::getInstance();
$DB = $registry['DB'];
$sql = "SELECT * FROM `task` ORDER BY task_name ASC";
$result = $DB->fetchAll($sql);
$page=$this->_getParam('page',1);
$paginator = Zend_Paginator::factory($result);
$paginator->setItemCountPerPage(3);
$paginator->setCurrentPageNumber($page);
$this->view->paginator=$paginator;
}

这是这个 Action 的 View

 <table border="1" align="center">
<tr>
<th>Task Name</th>
<th>Task Assign To</th>
<th>Action</th>
</tr>
<?php
foreach($this->paginator as $record){ ?>
<tr>
<td><?php echo $record['task_name']?></td>
<td><?php echo $record['task_assign']?></td>

<td>
<a href="edit/id/<?php echo $record['id']?>">Edit</a>
|
<a href="del/id/<?php echo $record['id']?>">Delete</a>
</td>
</tr>
<?php } ?>
</table>
<?php echo $this->paginationControl($this->paginator, 'Sliding', 'pagination.phtml'); ?>

这样做会给我这个错误

Fatal error: Cannot use object of type stdClass as array in C:\xampp\htdocs\zend_login\application\views\scripts\task\list.phtml on line 47

第47行指的是

 <td><?php echo $record['task_name']?></td>

我用这个函数将对象转换为数组

function objectToArray( $object ){
if( !is_object( $object ) && !is_array( $object ) ){
return $object;}
if( is_object( $object ) ){
$object = get_object_vars( $object );}
return array_map( 'objectToArray', $object );}
$paginator = objectToArray($this->paginator);

现在我在我的代码中更改它

foreach($paginator as $record){

还有这个

   <?php echo $this->paginationControl($paginator, 'Sliding', 'pagination.phtml'); ?>

但它现在给我这个错误

 Catchable fatal error: Argument 1 passed to Zend_View_Helper_PaginationControl::paginationControl() must be an instance of Zend_Paginator, array given in C:\xampp\htdocs\zend_login\library\Zend\View\Helper\PaginationControl.php on line 88

你能告诉我我错过了什么我是 zend 的新手所以请不要介意我尴尬的问题这是我的 ErrorController.php

class ErrorController extends Zend_Controller_Action{
public function errorAction()
{
$errors = $this->_getParam('error_handler');

switch ($errors->type) {
case Zend_Controller_Plugin_ErrorHandler::EXCEPTION_NO_ROUTE:
case Zend_Controller_Plugin_ErrorHandler::EXCEPTION_NO_CONTROLLER:
case Zend_Controller_Plugin_ErrorHandler::EXCEPTION_NO_ACTION:

// 404 error -- controller or action not found
$this->getResponse()->setHttpResponseCode(404);
$this->view->message = 'Page not found';
break;
default:
// application error
$this->getResponse()->setHttpResponseCode(500);
$this->view->message = 'Application error';
break;
}

// Log exception, if logger available
if ($log = $this->getLog()) {
$log->crit($this->view->message, $errors->exception);
}

// conditionally display exceptions
if ($this->getInvokeArg('displayExceptions') == true) {
$this->view->exception = $errors->exception;
}

$this->view->request = $errors->request;
}

public function getLog()
{
$bootstrap = $this->getInvokeArg('bootstrap');
if (!$bootstrap->hasPluginResource('Log')) {
return false;
}
$log = $bootstrap->getResource('Log');
return $log;
}
}

最佳答案

你只需要将它作为一个对象来调用:

<td><?php echo $record->task_name; ?></td>

更新:聊天中解决的所有其他错误

关于zend-framework - 通过 Zend Paginator 分页给出错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8835773/

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