我的index.php看起来是这样的：PHP回声变量从包括之前包括
<!DOCTYPE html>
<html>
<head>
<title><?php echo $title?></title>
</head>
<body>
<p>Hi, this is my homepage. I'll include some text below.</p>
<?php
include "myfile.php";
?>
</body>
</html>
“myfile.php” 例如包括以下内容：
<?php
$title = "Hi, I'm the title tag…";
?>
<p>Here's some content on the site I included!</p>
输出的<p> GOTS但<title>标签保持空白。
如何从包含的网站获得$title并在<body>标签中显示包含的内容？
这里是我的解决方案：
index.php
<?php
// Turn on the output buffering so that nothing is printed when we include myfile.php
ob_start();
// The output from this line now go to the buffer
include "myfile.php";
// Get the content in buffer and turn off the output buffering
$body_content = ob_get_contents();
ob_end_clean();
?>
<!DOCTYPE html>
<html>
<head>
<title><?php echo $title?></title>
</head>
<body>
<p>Hi, this is my homepage. I'll include some text below.</p>
<?php
echo $body_content; // Now, we have the output of myfile.php in a variable, what on earth will prevent us from echoing it?
?>
</body>
</html>
myfile.php保持不变。
P.S：由于诺曼建议，可以先include 'myfile.php';在index.php的开头，然后身体标记内echo $content;，并改变myfile.php弄成这个样子：
<?php
$title = "Hi, I'm the title tag"
ob_start();
?>
<p>Here's some content on the site I included!</p>
<?php
$content = ob_get_contents();
ob_end_clean();
?>
还有另一种方式来做到这一点：
的index.php
<?php
include "myfile.php";
?>
<!DOCTYPE html>
<html>
<head>
<title><?php echo $title?></title>
</head>
<body>
<p><?php $content; ?></p>
</body>
</html>
myfile.php
<?php
$title = "Hi, I'm the title tag…";
$content = "Here's some content on the site I included!";
?>
你使用echo $标题屏幕显示。 –
你想做什么。 'include'用于包含文件。让你的问题更清楚一点。 – Sasikumar
在使用'$ title'变量之前，您的文件必须包含在内。 – Sasikumar