在 PHP 中渲染视图
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14143865/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Render a view in PHP
提问by prometheus
I am writing my own MVC framework and has come to the view renderer. I am setting vars in my controller to a View object and then access vars by echo $this->myvar in the .phtml script.
我正在编写我自己的 MVC 框架并且已经来到了视图渲染器。我将控制器中的 vars 设置为 View 对象，然后通过 echo $this->myvar 在 .phtml 脚本中访问 vars。
In my default.phtml I call the method $this->content() to output the viewscript.
在我的 default.phtml 中，我调用方法 $this->content() 来输出视图脚本。
This is the way I do it now. Is this a proper way to do that?
这就是我现在的做法。这是一个正确的方法吗？
class View extends Object {
protected $_front;
public function __construct(Front $front) {
$this->_front = $front;
}
public function render() {
ob_start();
require APPLICATION_PATH . '/layouts/default.phtml' ;
ob_end_flush();
}
public function content() {
require APPLICATION_PATH . '/views/' . $this->_front->getControllerName() . '/' . $this->_front->getActionName() . '.phtml' ;
}
}
回答by Martin Samson
Example of a simple view class. Really similar to yours and David Ericsson's.
简单视图类的示例。与您和大卫爱立信的非常相似。
<?php
/**
* View-specific wrapper.
* Limits the accessible scope available to templates.
*/
class View{
/**
* Template being rendered.
*/
protected $template = null;
/**
* Initialize a new view context.
*/
public function __construct($template) {
$this->template = $template;
}
/**
* Safely escape/encode the provided data.
*/
public function h($data) {
return htmlspecialchars((string) $data, ENT_QUOTES, 'UTF-8');
}
/**
* Render the template, returning it's content.
* @param array $data Data made available to the view.
* @return string The rendered template.
*/
public function render(Array $data) {
extract($data);
ob_start();
include( APP_PATH . DIRECTORY_SEPARATOR . $this->template);
$content = ob_get_contents();
ob_end_clean();
return $content;
}
}
?>
Functions defined in the class will be accessible within the view like this:
类中定义的函数可以在视图中访问，如下所示：
<?php echo $this->h('Hello World'); ?>
回答by David Ericsson
Here's an example of how i did it :
这是我如何做到的一个例子：
<?php
class View
{
private $data = array();
private $render = FALSE;
public function __construct($template)
{
try {
$file = ROOT . '/templates/' . strtolower($template) . '.php';
if (file_exists($file)) {
$this->render = $file;
} else {
throw new customException('Template ' . $template . ' not found!');
}
}
catch (customException $e) {
echo $e->errorMessage();
}
}
public function assign($variable, $value)
{
$this->data[$variable] = $value;
}
public function __destruct()
{
extract($this->data);
include($this->render);
}
}
?>
I use the assign function from out my controller to assign variables, and in the destructor i extract that array to make them local variables in the view.
我使用我的控制器中的分配函数来分配变量，并在析构函数中提取该数组以使它们成为视图中的局部变量。
Feel free to use this if you want, i hope it gives you an idea on how you can do it
如果您愿意，请随意使用它，我希望它能让您了解如何做到这一点
Here's a full example :
这是一个完整的例子：
class Something extends Controller
{
public function index ()
{
$view = new view('templatefile');
$view->assign('variablename', 'variable content');
}
}
And in your view file :
在您的视图文件中：
<?php echo $variablename; ?>
