icE1
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[quote author=eipihipl link=topic=108011.msg434185#msg434185 date=1158242881] you've forgotten the content-type HTTP header header('content-type: image/jpeg'); cheers [/quote] Sorry but which part of my code should I insert this code in?
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Hi, I'm trying to retrieve an image that I stored in a mySQL database. I'm able to get the name, id but however unable to get the image to show up. The image shows up as: ótvwnY·ÌôMëÉØö,`âJ5œ››úõÐ׫hÚnÑ'«©”Œèè褤£üüüÎÎÎæ¶y‰ŽØ’æææN…QÔÔÔŠ™¼EÛúúúF¶üÊñ…u’;¹s§øÚÚÚòiVò“…–––q†nkÝÝÞQºººNm±°°°”jvdùèåááá{“Å6«õxf/W®‘¤Í¶¶¶ÚåøÑÑÑd—ø°¸ÈúârÎɹ½½¾êÑjòòòø¸sÈÈÈööö™‡vïïðMXnÅÅÅ=Âøøø¨†vNHÆ€€©©©ôôô‹SLñõýíí튋 My code for image retrieve is as follow: [code]<html> <head> <title>Download File From MySQL</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> </head> <body> <?php include 'config.php'; include 'opendb.php'; $query = "SELECT id, name, content 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, $content) = mysql_fetch_array($result)) { ?> <?php echo $name; echo $id; echo $content; ?> <?php } } include 'closedb.php'; ?> </body> </html> <?php if(isset($_GET['id'])) { // if id is set then get the file with the id from database include 'config.php'; include 'opendb.php'; $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; include 'closedb.php'; exit; } ?>[/code] Can anyone help me? What have I left out?