我的index.php看起來是這樣的：PHP回聲變量從包括之前包括
<!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」 例如包括以下內容：
<?php
$title = "Hi, I'm the title tag…";
?>
<p>Here's some content on the site I included!</p>
輸出的<p> GOTS但<title>標籤保持空白。
如何從包含的網站獲得$title並在<body>標籤中顯示包含的內容？
這裏是我的解決方案：
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保持不變。
P.S：由於諾曼建議，可以先include 'myfile.php';在index.php的開頭，然後身體標記內echo $content;，並改變myfile.php弄成這個樣子：
<?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();
?>
還有另一種方式來做到這一點：
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 $標題屏幕顯示。 –
你想做什麼。 'include'用於包含文件。讓你的問題更清楚一點。 – Sasikumar
在使用'$ title'變量之前，您的文件必須包含在內。 – Sasikumar