最简单/最快捷的“渲染” PHP视图文件方式
[英]Simplest/quickest way of 'rendering' view file in PHP
问题描述
Lets say i have a simple view file called MyView.php: 
可以说我有一个名为MyView.php的简单视图文件：
<!DOCTYPE html>
<html>
<head>
<title><?=$title?></title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
</head>
<body>
<?=$cotent?>
</body>
</html>
And i have a method in my code called render() : 我的代码中有一个方法称为render() ：
<?
function render($data)
{
}
?>
I need to call MyView.php from there and pass $title and $cotent variables to it. 我需要从那里调用MyView.php并将$title和$cotent变量传递给它。
I know that there is a way of doing that by just replacing <title><?=$title?></title> to, lets say, <title>{TITLE}</title> and then in render($data) just load MyView.php into variable and with preg_replace() replace all {BLAH} with $blah . 我知道有一种方法可以通过将<title><?=$title?></title>替换为<title>{TITLE}</title>然后在render($data)替换。只需将MyView.php加载到变量中，并使用preg_replace()将所有{BLAH}替换$blah 。
Is there any other simple way of doing this? 还有其他简单的方法吗？
Frameworks shall not be used. 不得使用框架。 Everything from scratch. 从头开始的一切。
1 个解决方案
解决方案1
3 已采纳 2013-03-13 22:07:58
Yes. 是。 From my personal library: 从我的个人图书馆：
function renderTemplate($tmpl, $__vars=array()) {
extract($__vars, EXTR_SKIP);
include($tmpl);
}
renderTemplate("MyView.php", array( "title" => "My Title", "content" => "My Content" ));
If you wanted to render to a string, you could modify it a bit: 如果要呈现为字符串，则可以对其进行一些修改：
function renderTemplateToString($tmpl, $__vars=array()) {
ob_start();
extract($__vars, EXTR_SKIP);
include($tmpl);
return ob_get_clean();
}
Note that renderTemplate() needs to be kept in its own function, even if you're only calling it once: it's using the function's variable scope to keep template variables separate from other variables. 请注意，即使只调用一次renderTemplate()需要保留在其自己的函数中：它使用函数的变量作用域将模板变量与其他变量分开。
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