我试图从模块到项目基础视图渲染视图,但它给出了错误.
我尝试了以下组合没有任何运气.它给出错误"DefaultController无法找到所请求的视图"appsMenu"."
echo $this->renderPartial("appsMenu",array("moduleName"=>""),true, true);
echo $this->renderPartial("//appsMenu",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("/protected/views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("/views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("//protected/views/site/appsMenu",array("moduleName"=>""));
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并尝试使用扩展
echo $this->renderPartial("appsMenu.php",array("moduleName"=>""),true, true);
echo $this->renderPartial("//appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("/protected/views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("/views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("//protected/views/site/appsMenu.php",array("moduleName"=>""));
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我在"表单"模块中并尝试呈现文件"protected/views/site/appsMenu.php".Plz帮帮我..
在backend/config/main文件中,有一个对模块类的引用:
'modules' => [
'cropk' => [
'class' => 'app\modules\cropk\CropK',
]
],
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在vendor/xxx/cropk目录中,有以下类CropK:
namespace app\modules\cropk;
class CropK extends \yii\base\Module {
public function init() {
parent::init();
}
}
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的供应商/ XXX/cropk /控制器/ DefaultController:
namespace app\modules\cropk\controllers;
use yii\web\Controller;
class DefaultController extends Controller {
public function actionIndex() {
return $this->render('index');
}
}
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但是当我访问URL http://admin.cropk.dev/cropk时,我收到此错误:
类app\modules\cropk\CropK不存在
我不能把模块放在后端目录之外吗?我怎样才能做到这一点?
我在以下结构的模块中有一个图像:vendor/myvendorname/mymodulename/assets/img/delete-icon.png
我需要<img>通过JavaScript向页面添加一个,并且它可能具有src指向该属性的属性delete-icon.png.
$("#delete").attr("src", "?");
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如果将图像放入Yii创建的资产目录中,如何引用该图像?获得这条道路的方法是什么?
在我的Yii应用程序中,我有一个名为admin的模块,因此模块类是AdminModule,它扩展了CWebModule类,位于AdminModule.php文件中.根据文档,CWebModule具有layout在所有模块控制器之间共享的属性,以防控制器本身没有定义任何布局.
我的控制器没有定义任何布局,在AdminModule.php我把它:
$this->layout='webroot.themes.bootstrap.views.layouts.column2';
$this->layoutPath = Yii::getPathOfAlias('webroot.themes.bootstrap.views.layouts');
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但是,管理模块中的控制器仍在使用其他布局,我认为它是在components目录中的Controller.php中定义的.这是为什么?如何为特定模块设置共享布局?