[英]Layout no render on some actions in Zend Expressive?
Is it possible to set layout no render in one Action (or set of Actions)? 是否可以在一个Action(或一组Actions)中设置布局无渲染?
As I know I can set default layout in config, which will render on every page. 据我所知,我可以在config中设置默认布局,它将在每个页面上呈现。 I can change it in Action bay passing the 'layout' variable with value, but is it possible to not render layout at all? 我可以在Action中更改它,并使用值传递'layout'变量,但是有可能根本不渲染布局吗?
class IndexAction
{
private $template;
public function __construct(Template $template){ ... }
public function __invoke($request, $response, $next = null)
{
if(!$request->hasHeader('X-Requested-With')){
$data = ['layout' => 'new\layout']; //change default layout to new one
}
else{
$data = ['layout' => false]; //I need only to return view ?
}
return new HtmlResponse($this->template->render(
'web::index', $data
));
}
}
Actions and templates in Zend Expressive are one-to-one, so I think that the decision whether layout should be rendered or not should be made in the appropriate template itself. Zend Expressive中的动作和模板是一对一的,所以我认为应该在适当的模板本身中决定是否应该呈现布局。 Basically, it's a matter of omitting <?php $this->layout('layout::default'); ?>
基本上,这是省略<?php $this->layout('layout::default'); ?>
<?php $this->layout('layout::default'); ?>
from your action template. <?php $this->layout('layout::default'); ?>
来自您的操作模板。
In your particular example, that condition you have should result in choosing different template to be rendered (one that doesn't include layout), instead of a solution with sending a flag to the template. 在您的特定示例中,您所具有的条件应该导致选择要呈现的不同模板(不包括布局的模板),而不是向模板发送标志的解决方案。 For example: 例如:
$templateName = $request->hasHeader('X-Requested-With')
? 'template-without-layout'
: 'template-with-layout';
return new HtmlResponse($this->template->render($templateName));
Now it's available, just set layout = false 现在它可用,只需设置layout = false
return new HtmlResponse($this->template->render(
'web::index',
['layout' => false]
));
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.