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Zend框架控制器操作助手
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Stack Overflow用户
提问于 2010-06-09 10:02:22
回答 1查看 3.4K关注 0票数 1

在添加action类之后,我得到了致命错误。我正在尝试加载与所谓布局相对应的布局。下面是我的代码片段:

首先,我在应用程序/控制器/帮助程序中添加了一个助手类:

代码语言:javascript
复制
class Zend_Controller_Action_Helper_Layout extends Zend_Controller_Action_Helper_Abstract {

public $pluginLoader;


public function __construct() 
{
    // TODO Auto-generated Constructor
    $this->pluginLoader = new Zend_Loader_PluginLoader ();
}


public function preDispatch()
{
        $bootstrap = $this->getActionController()->getInvokeArg('bootstrap');
        $config = $bootstrap->getOptions();
        $module = $this->getRequest()->getModuleName();
        if (isset($config[$module]['resources']['layout']['layout'])) {
            $layoutScript = $config[$module]['resources']['layout']['layout'];
            $this->getActionController()->getHelper('layout')->setLayout($layoutScript);
        }

    }

}

然后我在bootstrap.php:中添加了一个加载程序

代码语言:javascript
复制
protected function _initLayoutHelper() {

    $this->bootstrap('frontController');
    Zend_Controller_Action_HelperBroker::addPath(APPLICATION_PATH .'/controllers/helpers');

    $layout = Zend_Controller_Action_HelperBroker::addHelper(new Zend_Controller_Action_Helper_Layout());
}

以下是我的application.ini:

代码语言:javascript
复制
[production]
autoloaderNamespaces.tree = "Tree_"
phpSettings.display_startup_errors = 0
phpSettings.display_errors = 0
includePaths.library = APPLICATION_PATH "/../library"
bootstrap.path = APPLICATION_PATH "/Bootstrap.php"
bootstrap.class = "Bootstrap"
resources.frontController.controllerDirectory = APPLICATION_PATH "/controllers"
resources.frontController.moduleDirectory = APPLICATION_PATH "/modules"
resources.frontController.helperDirectory = APPLICATION_PATH "/controllers/helpers"

resources.modules[] = ""
contact.resources.frontController.defaultControllerName = "index"
resources.layout.layoutPath = APPLICATION_PATH "/layouts/scripts"
resources.layout.layout = layout
admin.resources.layout.layout = admin
#admin.resources.layout.layoutPath = APPLICATION_PATH "/layouts/scripts"

resources.view[] =

[staging : production]

[testing : production]
phpSettings.display_startup_errors = 1
phpSettings.display_errors = 1

[development : production]
phpSettings.display_startup_errors = 1
phpSettings.display_errors = 1

运行这段代码时,我收到以下错误:

警告: include(Zend\Controller\Action\Helper\LayoutLoader.php) function.include:未能打开流:在第83行的D:\personal\proj\renovate\library\Zend\Loader.php中没有这样的文件或目录

警告:'Zend\Controller\Action\Helper\LayoutLoader.php‘() function.include:在第83行的D:\personal\proj\renovate\library\Zend\Loader.php中打开包含(include_path='D:\personal\proj\renovate\application/../library;D:\personal\proj\renovate\library;.;C:\php5\pear')的D:\personal\proj\renovate\library\Zend\Loader.php失败

致命错误:在第33行的D:\personal\proj\renovate\application\Bootstrap.php中找不到类'Zend_Controller_Action_Helper_LayoutLoader‘

请让我知道,我怎样才能走出这个问题。我是的初学者。

EN

回答 1

Stack Overflow用户

发布于 2010-06-09 11:52:04

我认为包含路径是有效的。以下是我的index.php

定义应用程序目录的路径

代码语言:javascript
复制
defined('APPLICATION_PATH') || define('APPLICATION_PATH', realpath(dirname(__FILE__) . '/../application'));

定义应用环境

代码语言:javascript
复制
defined('APPLICATION_ENV') || define('APPLICATION_ENV', (getenv('APPLICATION_ENV') ? getenv('APPLICATION_ENV') : 'development'));

确保library/include_path

代码语言:javascript
复制
set_include_path(implode(PATH_SEPARATOR, array(realpath(APPLICATION_PATH . '/../library'),get_include_path(),)));

/** Zend_Application */
require_once 'Zend/Application.php'; 

// Create application, bootstrap, and run
$application = new Zend_Application(APPLICATION_ENV, APPLICATION_PATH . '/configs/application.ini');
$application->bootstrap()->run();
票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/3004778

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