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Editorial Team
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Asked: 2026-05-26T16:22:57+00:00

Here’s the code: <?php // For use in creating individual page $tpl_file = submission.php;

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Here’s the code:

<?php
// For use in creating individual page
$tpl_file = "submission.php";
$tpl_path = "templates/";
$submissions_path = "submissions/";



// For use in querying submitter name

$username = $_GET['username'];
session_start();
$_SESSION['username'] = $username; 

//Database Information

$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";

//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());


$name = $_POST['name'];
$filename = $_POST['filename'];   
$submitter = $username;
list($width, $height) = getimagesize("$filename");
$type = exif_imagetype($_POST['filename']);

$checkuser = mysql_query("SELECT filename FROM images WHERE filename='$filename'");

$filename_exist = mysql_num_rows($checkuser);

if($filename_exist > 0){
    echo "I'm sorry but this image has already been submitted. Please feel free to try another.";
    unset($filename);
    include 'upload.php';
    exit();
}


if (exif_imagetype($_POST['filename']) == IMAGETYPE_GIF) {
    echo "Sorry, but we can't accept GIFs. Please feel free to try uploading another.";
    unset($filename);
    include 'upload.php';
    exit();
}



$query = "INSERT INTO images (name, filename, submitter, width, height, type)
VALUES('$name', '$filename', '$submitter', '$width', '$height', $type)";
mysql_query($query) or die(mysql_error());
mysql_close();

echo "Thanks for your submission!<br/> Upload another <a href='/~lyons/upload.php'>here</a>!";

$tpl = file_get_contents($tpl_path.$tpl_file);
$php_file_name = $name.".php";

$fh = fopen($submissions_path.$php_file_name, "w");
fwrite($fh, $tpl);
fclose($fp); 
?>

When a user submits a picture, it is supposed to automatically create a page based on a template. Here’s the code for the template:

<html>
<title><?php echo $name; ?></title>
<head>
</head>

<body>
<h1><?php echo $name ?></h1>
Posted by: <?php echo $username ?>
<br/>
<img src="<?php echo $filename ?>"/>

</body>
</html>

As you might have already guessed, I want it to put in values for name, username, and filename that were derived in the first script where they submit the picture. However, it seems they don’t carry over. The page is created, but where ever it’s supposed to echo the values for the variables, it is blank. How can I include the values for those variables that I want to use in the created page?

Thanks in advance to whoever can help me.

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1 Answer

  1. Editorial Team

    I would suggest using a string like %name%, %username% etc. to mark placeholders for variables.

    Then, before writing to the file, try something like this:

    $tpl = preg_replace("(%([a-z_][a-z0-9_]*)%)ie",'$$1',$tpl);

    This will find, for example, %filename% and replace it with the contents of the variable $filename.

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