13

我正在尝试Slim php 框架

Slim 中是否可以有布局或子视图？我想将视图文件用作模板，并将变量用作单独加载的其他视图的占位符。

我该怎么做？

4

7 回答 7

21

文件名：myview.php

<?php
class myview extends Slim_View
{
    static protected $_layout = NULL;
    public static function set_layout($layout=NULL)
    {
        self::$_layout = $layout;
    }
    public function render( $template ) {
        extract($this->data);
        $templatePath = $this->getTemplatesDirectory() . '/' . ltrim($template, '/');
        if ( !file_exists($templatePath) ) {
            throw new RuntimeException('View cannot render template `' . $templatePath . '`. Template does not exist.');
        }
        ob_start();
        require $templatePath;
        $html = ob_get_clean();
        return $this->_render_layout($html);
    }
    public function _render_layout($_html)
    {
        if(self::$_layout !== NULL)
        {
            $layout_path = $this->getTemplatesDirectory() . '/' . ltrim(self::$_layout, '/');
            if ( !file_exists($layout_path) ) {
                throw new RuntimeException('View cannot render layout `' . $layout_path . '`. Layout does not exist.');
            }
            ob_start();
            require $layout_path;
            $_html = ob_get_clean();
        }
        return $_html;
    }

}
?>

例如 index.php：

<?php
require 'Slim/Slim.php';
require 'myview.php';

// instantiate my custom view
$myview = new myview();

// seems you need to specify during construction
$app = new Slim(array('view' => $myview));

// specify the a default layout
myview::set_layout('default_layout.php');

$app->get('/', function() use ($app) {
    // you can override the layout for a particular route
    // myview::set_layout('index_layout.php');
    $app->render('index.php',array());
});

$app->run();
?>

default_layout.php：

<html>
    <head>
        <title>My Title</title>
    </head>
    <body>
        <!-- $_html contains the output from the view -->
        <?= $_html ?>
    </body>
</html>
于 2011-10-29T03:18:51.003 回答
8

slim 框架使用了其他模板引擎，例如Twig或Smarty，因此只要您选择允许子视图的模板引擎，它就可以工作。有关 Slim 中的视图模板的更多信息，请查看此处。

于 2011-09-04T23:19:19.007 回答
7

我正在使用我的视图：

class View extends \Slim\View
{
    protected $layout;

    public function setLayout($layout)
    {
        $this->layout = $layout;
    }

    public function render($template)
    {
        if ($this->layout){
            $content =  parent::render($template);
            $this->setData(array('content' => $content));
            return parent::render($this->layout);
        } else {
            return parent::render($template);
        }
    }
}

您可以添加一些方法，例如setLayoutData(..)或appendLayoutData(..)

几分钟前：

class View extends \Slim\View
{

    /** @var string */
    protected $layout;

    /** @var array */
    protected $layoutData = array();

    /**
     * @param string $layout Pathname of layout script
     */
    public function setLayout($layout)
    {
        $this->layout = $layout;
    }

    /**
     * @param array $data
     * @throws \InvalidArgumentException
     */
    public function setLayoutData($data)
    {
        if (!is_array($data)) {
            throw new \InvalidArgumentException('Cannot append view data. Expected array argument.');
        }

        $this->layoutData = $data;
    }

    /**
     * @param array $data
     * @throws \InvalidArgumentException
     */
    public function appendLayoutData($data)
    {
        if (!is_array($data)) {
            throw new \InvalidArgumentException('Cannot append view data. Expected array argument.');
        }

        $this->layoutData = array_merge($this->layoutData, $data);
    }

    /**
     * Render template
     *
     * @param  string $template Pathname of template file relative to templates directory
     * @return string
     */
    public function render($template)
    {
        if ($this->layout){
            $content = parent::render($template);

            $this->appendLayoutData(array('content' => $content));
            $this->data = $this->layoutData;

            $template = $this->layout;
            $this->layout = null; // allows correct partial render in view, like "<?php echo $this->render('path to parial view script'); ?>"

            return parent::render($template);;
        } else {
            return parent::render($template);
        }
    }
}
于 2013-04-06T21:24:51.800 回答
4

搭配超薄 3

namespace Slim;
use Psr\Http\Message\ResponseInterface;

class View extends Views\PhpRenderer
{
    protected $layout;

    public function setLayout($layout)
    {
        $this->layout = $layout;
    }

    public function render(ResponseInterface $response, $template, array $data = [])
    {
        if ($this->layout){
            $viewOutput = $this->fetch($template, $data);
            $layoutOutput = $this->fetch($this->layout, array('content' => $viewOutput));
            $response->getBody()->write($layoutOutput);
        } else {
            $output = parent::render($response, $template, $data);
            $response->getBody()->write($output);
        }
        return $response;
    }
}

在布局中：

<?=$data['content'];?>
于 2016-05-11T12:49:48.907 回答
1

您可以在父视图中使用它：

$app = \Slim\Slim::getInstance();
$app->render('subview.php');
于 2014-03-31T03:18:58.717 回答
1

如果要在变量中捕获输出，就像使用视图的fetch()方法一样简单：

//assuming $app is an instance of \Slim\Slim
$app->view()->fetch( 'my_template.php', array( 'key' => $value ) );
于 2014-10-29T14:14:31.300 回答
0

对的，这是可能的。如果您不想使用任何模板引擎，例如 smarty 或 twig。如果您只想使用 .php 视图文件。按着这些次序。

步骤 1. 在项目的根目录中创建一个新目录。例如模板。

步骤 2. 打开 dependencies.php 并粘贴以下代码

$container = $sh_app->getContainer();

$container['view'] = function ($c) {
   $view = new Slim\Views\PhpRenderer('src/templates/');
   return $view;
};

第 3 步。在您的控制器的渲染视图代码的方法应该是这样的。

$this->container->view->render($response,'students/index.php',['students' => $data]);
于 2017-05-29T08:58:24.623 回答