Please Help - Download Script

[Anon]

I have an upload script which stores files on a specified directory on my C: drive and then stores the location and filename inside the DB which works with no problems - The insert statement looks like this

$sql = \"INSERT INTO coursework_table (fileloc, filename, filesize, filetype, user_id, upload_id) VALUES ( \'$fileloc\', \'$file_name\', \'$file_size\', \'$file_type\', \'$user_id\', \'$upload_id\') \";
$result=mysql_query(\"$sql\");

Fileloc is defined as $fileloc = \"/files/work/$file_name\";

I am confused on how to create a download script to download an uploaded file from my server based upon the upload_id.

I am a newbie, so take it easy on me :-)

At the moment, when i click on \'download\', the filename and path are displayed in the browser! and the file obviously does not dowload.

I have tried out the following script, but i get no joy, can anybody please help?

$query = \"select fileloc, filetype from upload_table where upload_id=\'$upload_id\' \";
$result = mysql_query(\"$query\");

$data = mysql_result($result,0,\"fileloc\");
$type = mysql_result($result,0,\"filetype\");

header ( \"Content-type: $type\");
echo $data;

[thefury]

$data = mysql_result($result,0,"fileloc");
$type = mysql_result($result,0,"filetype");

should be

$data = mysql_result($result,0);
$type = mysql_result($result,1);

[Anon]

Hi I have tried that but now get the follwing error message..

Warning: Unable to jump to row 1 on MySQL result index 2 in c:\program files\apache group\apache\htdocs\sis\primary\download.php on line 11

Warning: Cannot add header information - headers already sent by (output started at c:\program files\apache group\apache\htdocs\sis\primary\download.php:11) in c:\program files\apache group\apache\htdocs\sis\primary\download.php on line 16
/files/work/abireef.txt

My Download script currently looks like this now:

<?php include("dbconnect.php"); ?>
<?php

$query = "select fileloc, filetype from coursework_table where coursework_id='$coursework_id' ";
$result = mysql_query("$query");

$data = mysql_result($result,0);
$type = mysql_result($result,1);

header ( "Content-type: $type");
echo $data;

?>

Any suggestions?

-Abi