PHP 在包含之前回显来自包含的变量

Question

我的 index.php 看起来像这样：

<!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>

<p> 得到输出，但 <title> 标签保持空白。

我怎样才能从包含的站点获取 $title 并在 <body> 标签内显示包含的内容？

Answer 1

还有另一种方法：

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!";
?>

Answer 2

这是我的解决方案：

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不变。

P.S：按照Noman的建议，你可以先在index.php的开头include 'myfile.php';，然后在body标签里面echo $content;，然后改成myfile.php变成这样的东西：

<?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();
?>

PHP 在包含之前回显来自包含的变量

PHP echo variable from include before include

html

php

title

include