Posted by kenoli on 01/09/07 07:27

I am trying to retrieve an image stored in a blob field in a database
and display it in an html document. I call it into the file where I
want to display it using the following code:

<?php echo '<img src="echo_image.php" title="Personal Thumbnail."
alt="No image available.">';?>

The file echo_image.php contains the following code:

<?php
$tid = '18';

// Connect to the database
require_once ('../includes/scripts/mysql_connect_bh.php');

// Create the query
$sql = "SELECT image FROM tbl_person WHERE person_id=$tid";

// Query the database
$result = mysql_query("$sql") or die("Invalid query: " .
mysql_error());

// Send header
header("Content-type: image");

// Echo the image
echo mysql_result($result, 0);

?>

This works fine. The image is placed in my original document,

The problem is that I want send an id to the echo_image.php file so I
can tell it which record to select the image from. When I set a
session variable in the first file and try to retrieve it in the second
file like this:

<?php
session_start();

$_SESSION['id'] = 18;

php echo '<img src="echo_image.php" title="Personal Thumbnail."
alt="No image available.">';
?>

<?php
$tid = $_SESSION['id'];

// Connect to the database
require_once ('../includes/scripts/mysql_connect_bh.php');

// Create the query
$sql = "SELECT image FROM tbl_person WHERE person_id=$tid";

// Query the database
$result = mysql_query("$sql") or die("Invalid query: " .
mysql_error());

// Send header
header("Content-type: image");

// Echo the image
echo mysql_result($result, 0);

?>

I get an error saying "no image available.

How can I send an id value from the first file to the second so that I
can indicate which record to retrieve the image from?

Is there some other way to do this?

Also, why doesn't the following code work in the second file. When I
do this, the image data gets returned rather than the interpreted
image.

<?php
require_once ('../includes/scripts/mysql_connect_bh.php');

$tid = '18';

// Create the query
$sql = "SELECT image FROM tbl_person WHERE person_id=$tid";

// Query the database
$result = mysql_query("$sql") or die("Invalid query: " .
mysql_error());

$row = mysql_fetch_array($result, MYSQL_ASSOC);

// Send header
header("Content-type: image");
echo $row['image'];

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