Solved: display images in php

facing problem in displaying the images that are stored in databases the following code is to display the images

<?php

$link = mysql_connect('localhost', 'root', '');
if (!$link) {
die('Not connected : ' . mysql_error());
}

$db_selected = mysql_select_db('upload', $link);

if (!$db_selected) {
die ('Database error : ' . mysql_error());
}

$sql = "SELECT * FROM images";

header("Content-type: image/jpeg");
echo mysql_result($result, 0,'imageData');

?>
the following code is to upload the images in the database
<?php

$link = mysql_connect('localhost', 'root', '');
if (!$link) {
die('Not connected : ' . mysql_error());
}

$db_selected = mysql_select_db('upload', $link);
if (!$db_selected) {
die ('Database error : ' . mysql_error());
}

$maxFileSize = "1000000";

$image_array            = array("image/jpeg","image/jpg","image/gif","image/bmp","image/pjpeg","image/png");

$fileType = $_FILES['userfile']['type'];

$msg = '';

if(@$_POST['Submit'])
{

if (in_array($fileType, $image_array))
{

if(is_uploaded_file($_FILES['userfile']['tmp_name']))
{

if($_FILES['userfile']['size'] < $maxFileSize)
{

$imageData =addslashes (file_get_contents($_FILES['userfile']['tmp_name']));

$sql = "INSERT INTO images ( imageData) VALUES ('$imageData')";

mysql_query($sql) or die(mysql_error());
            $msg = " Data successfully uploaded";
}
else
       {

       $msg = ' Error : File size exceeds the maximum limit ';

       }
}
}
else
{
$msg = 'Error: Not a valid image ';
}

}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html>
<body>
       <span style="color:#FF0000"><?php echo $msg; ?></span><br />

<h3>Select File to upload</h3>

<form enctype="multipart/form-data" action="" method="post">
<input name="userfile" type="file" />
<input type="submit" value="Submit" name="Submit"/>
</form>
</body>
</html>
i am able to upload the images here in the database but i am unable to display all the images that are stored in the database.help me out experts

$db_selected = mysql_select_db('upload', $link); if (!$db_selected) { die ('Database error : ' . mysql_error()); } $sql = "SELECT * FROM images"; $result = mysql_query($sql); $row = mysql_fetch_array($result); header("Content-type: image/jpeg"); echo $row['imageData']; ?>