1

aplication/model/newsModel.php中的模型

class models_newsModel extends Zend_Db_Table_Abstract{
}

application\controllers\LoginController.php中的控制器

我尝试创建对象来建模

$this->news = new models_newsModel();

并得到错误：*Class 'models_newsModel' not found in C:\wamp\www\sistema\application\controllers\LoginController.php on line 27*

也许问题出在我的 application.ini 中

[production]
phpSettings.display_startup_errors = 1
phpSettings.display_errors = 1
bootstrap.path = APPLICATION_PATH "/Bootstrap.php"
bootstrap.class = "Bootstrap"
resources.frontController.controllerDirectory = APPLICATION_PATH "/controllers"
; ADD THE FOLLOWING LINES
;resources.layout.layout = "layout"
;resources.layout.layoutPath = APPLICATION_PATH "/layouts/scripts"

[staging : production]

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

[development : production]
phpSettings.display_startup_errors = 1
phpSettings.display_errors = 1
4

1 回答 1

1

您是否创建了资源加载器？您可以在引导程序中对其进行初始化：

protected function _initAppAutoload()
{
    $resourceLoader = new Zend_Loader_Autoloader_Resource(array(
        'namespace' => '',
        'basePath'  => APPLICATION_PATH));    
}

默认情况下，它希望模型存储在：下application/models/并命名为Model_*或Model_News在您的情况下。但是，如果您愿意，可以使用addResourceType方法更改模型位置和命名方案。在这里阅读更多

于 2013-07-23T02:22:43.283 回答