Question

I want to echo block of code which is dynamically generated. For example:

<?php
$cid = $camp_id;

$hostname = "$host";
$db_user = "$dbuser";
$db_pass = "$dbpass";
$db_name = "$dbname";

$mysqli = new mysqli();
$mysqli->connect($hostname, $db_user, $db_pass, $db_name);
if ($mysqli->connect_errno) {
    echo "Failed to connect to MySQL: " . $mysqli->connect_error;
}

etc....
?>

I got access to $camp_id and other variables because they are in the file which is included.
I tried to store this code in variable with < pre> and < code> tag and echo after that but couldn’t make it work.

Also how can I insert $camp_id to this. Below is example what I think (I know it’s not correct just for understanding.

$generated_code = "<.code><?php $cid = <?php echo $camp_id;?> $hostname = $host; etc... </code > ?>";

I used space and dot before code and pre because if not it doesn’t show as tag..

Thanks

Answer 1

You could also try like this:

<?php
    ob_start();
?>

    <code>$cid = <?php echo $camp_id; ?> , $hostname = <?php echo $host; ?></code>

<?php
    echo ob_get_clean();
?>

Depending on the circumstances and what your code is like, using ob_start() and ob_get_clean() functions allow your code to be more legible in color coded IDEs, since your output wont look like one solid block of color, instead it will be styled like it should in html for better readability.