PHP echo variable from include before include
Question
My index.php looks like this:
<!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" eg includes the following:
<?php
$title = "Hi, I'm the title tag…";
?>
<p>Here's some content on the site I included!</p>
The <p> gots outputted but the <title> tag stays blank.
How can I reach to get $title from the included site and also show the included content inside the <body> tag?
2 answers
solution1
1 ACCPTED 2016-09-19 07:55:49
Here is my solution:
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 is unchanged.
PS: As Noman suggested, you can first include 'myfile.php'; at the beginning of index.php , then echo $content; inside body tag, and change myfile.php into something like this:
<?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();
?>
solution2
0 2016-09-19 07:30:14
There is another way to do it:
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!";
?>
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