我尝试将视图从模块渲染到项目基础视图，但它给出了错误。
我尝试了下面的组合，但没有任何运气。它给出了错误“DefaultController找不到请求的视图“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"=>""));并尝试扩展
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 $this->renderPartial("//protected/views/site/appsMenu.php",array("moduleName"=>""));我在“表单”模块，并试图呈现一个文件“protected/views/site/appsMenu.php”。请帮助我。
3条答案
按热度按时间7xzttuei1#
使用
//：这可以在文档中看到
应用程序中的绝对视图：视图名称以双斜杠“//”开头。在这种情况下，将在应用程序的视图路径下搜索视图。此语法从1.1.3版开始可用。
kq4fsx7k2#
这个讨厌的小东西对我起了作用
使用
//作为$root/protected/views的别名，然后将../位放入其中，得到$root/protected/views/../modules/$m/views/$c/$v，实际上是$root/protected/modules/$m/views/$c/$v当然，为上面的$X和/或MyXXX值输入合理的值。
fzsnzjdm3#
如果您无法使用yii渲染文件，请使用上面的行