PHP5 - Image display from mysql using php

i want to display image from mysql using php.
i uploaded file to db corectly.
header("Content-Disposition: inline; filename=$trend_file_name"); i can get pdf file correctly,but can't get jpeg files.i don't identify the problem .......help me plz

What is it doing / not doing? Please be more specific.

A JPEG file of course, has the mime type "image/jpeg". Did you properly store the mime type and are using header() with the 'Content-type' header to output that? What about the file size ('Content-length')?

Are you storing the files in at least a MEDIUMBLOB type field?

Can you show us the relevant code?

Setting aside for the moment the fact that storing files in the database is inadvisable, the content type is the most important header to send in this case. In many cases cache control headers are also necessary. A good reference for this is the PHP manual page for header(), don't forget to read the comment section.

if((isset ($_FILES['trend_file']['name']) && is_uploaded_file($_FILES['trend_file']['tmp_name'])))
{
$fileName = $_FILES['trend_file']['name'];
$tmpName = $_FILES['trend_file']['tmp_name'];
$fileSize = $_FILES['trend_file']['size'];
$fileType = $_FILES['trend_file']['type'];
$fp = fopen($tmpName, 'r');
$content_file = fread($fp, filesize($tmpName));
$content_file = addslashes($content_file);
fclose($fp);
if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
}


$grpfil=mysql_query("update `property_trend` set `trend_file_size`='$fileSize',`trend_file_type`='$fileType',`trend_file_name`='$fileName',`trend_fil e`='$content_file' where `trend_id`='$filepathid'") or die("111");
}

i used for uploading image
**************************************************
$select_doc=mysql_query("select `trend_file_size`,`trend_file_type`,`trend_file_name`,`trend_file` from `property_trend` where `trend_id`='$trend_id' ");
$row = mysql_fetch_array($select_doc);


$trend_file=$row['trend_file'];
$trend_file_name=$row['trend_file_name'];

$size =$row['trend_file_size'];
$type=$row['trend_file_type'];

header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: inline; filename=$trend_file_name");

echo $trend_file;

******************************************' this code for dispaly ..

this the code am using. trend_file_type value is image/jpeg
and trend_file type is mediumblob ...............


Any hope?
i want to display this image on html page ,not for downloading....

How are you linking to this file in order to display the image?

<img src="getTrendImage.php?tend_id=4" />


Is it just trying to download the image instead of display it in the browser? If you go ahead and download it, does it open properly in an image program (ie is it a valid image)?

addslashes() is very insecure, you will probably need to base64 encode that data, but since I would never store such data in a database, I'm not certain of that. You also escape the data twice if magic_quotes is not enabled, which makes no sense.