Hi,
I have image paths stored inside a database, how do I create a link, in order to display the down dialog box to enable the user to download the file/image?
[code=php]<html> <head> <title>Download File From MySQL</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> </head> <body> <?php $con = mysql_connect("HOST","USERNAME","PASSWORD"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("DATABASE", $con); $query = "SELECT id, name FROM upload"; $result = mysql_query($query) or die('Error, query failed'); if(mysql_num_rows($result) == 0) { echo "Database is empty <br>"; } else { while(list($id, $name) = mysql_fetch_array($result)) { ?> <a href="download.php?id=<?php=$id;?>"><?php=$name;?></a> <br> <?php } } mysql_close($con); ?> </body> </html>[/code]
[code=php]$query = "SELECT id, name, field here FROM upload"; [/code]
[code=php]while(list($id, $name, $field here) = mysql_fetch_array($result))[/code]
[code=php]<?php if(isset($_GET['id'])) { // if id is set then get the file with the id from database $con = mysql_connect("HOST","USERNAME","PASSWORD"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("DATABASE", $con); $id = $_GET['id']; $query = "SELECT name, type, size, content " . "FROM upload WHERE id = '$id'"; $result = mysql_query($query) or die('Error, query failed'); list($name, $type, $size, $content) = mysql_fetch_array($result); header("Content-length: $size"); header("Content-type: $type"); header("Content-Disposition: attachment; filename=$name"); echo $content; mysql_close($conn); exit; } ?>[/code]
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