Displayig images from MySQL database

sharapov

I am trying to dispaly images from MySQL databse without showing the actual file name in HTML page. I have two pages index.php and view.php. View.php actually the one that displays the image, but its called from index.php file. The images should be shown by it's id but when the page loads image is not displayed. CAn somebody help me to find an error in my code? Below are index.php and view.php.

index.php:

<!DOCTYPE HTML PUBLIC
"-//W3C//DTD HTML 4.0 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">

<?php

/
db.inc stores some common functions to access MySQL DBMS.
/

include 'db.inc';

$query = "SELECT product_id, category_id, item_number, item_name, image FROM products";

// Connect to the MySQL DBMS
if (!($connection = @ mysql_pconnect($hostName,
$username,
$password)))
showerror();

if (!mysql_select_db("catalog", $connection))
showerror();

// Run the query on the DBMS
if (!($result = @ mysql_query ($query, $connection)))
showerror();
?>
<h1>Image database</h1>

<h3>Click <a href="insert.php">here</a> to upload an image.</h3>

<?php
require 'disclaimer';

if ($row = @ mysql_fetch_array($result))
{
?>

<table>
<col span="1" align="right">
<tr>
   <th>Item Number</th>
   <th>Item Name</th>
   <th>Image</th>

</tr>

<?php
do
{
?>
<tr>
<td><?php echo "{$row["item_number"]}";?></td>

<td><?php echo "{$row["item_name"]}";?></td>
<td><?php echo "<img src=\"view.php?file={$row["product_id"]}\">";?></td>

</tr>
<?php
} while ($row = @ mysql_fetch_array($result));
?>
</table>
<?php
} // if mysql_fetch_array()
else
echo "<h3>There are no images to display</h3>\n";
?>
</body>

view.php:

<?php
/
db.inc stores some common functions to access MySQL DBMS.
/

include 'db.inc';

$file = clean($file, 4);

if (empty($file))
exit;

// Connect to the MySQL DBMS
if (!($connection = @ mysql_pconnect($hostName,
$username,
$password)))
showerror();

if (!mysql_select_db("catalog", $connection))
showerror();

$query = "SELECT image FROM products
WHERE product_id = $file";

// Run the query on the DBMS
if (!($result = @ mysql_query ($query,$connection)))
showerror();

$data = @ mysql_fetch_array($result);

if (!empty($data["image"]))
{
// Output the MIME header
//header("Content-Type: {$data["mimeType"]}");
// Output the image
echo $data["image"];
}
?>

Anon

I don't know if this'll help or not, because it doesn't look like its along the same line as what your code is, but:

I needed to be able to display thumbnails, so I:

Had a list of file names (pics) in a directory.

I recreated that list in a mysql table/column(along with a bunch of other specs about the pics in seperate columns).

to make a long story short; folks could query for required files(pics).

Put the return in to an array and loop:

while ($row=mysql_fetch_array($result, MYSQL_NUM))
{
$record=$row[0];

//and list them out on to one page:

       echo ("$record");
    }

You can wrap the 'echo' statement in an <a href""> so that the individual images act as links

theres some other formating stuff to go along with this so that they thumbnail correctly, but maybe this'll help out a bit.