Hi,
your code seems fine except for the $result_array["speaker_id"] variable used when
calling the image files.
Shouldn't this be $result_array3["speaker_id"] and $result_array4["speaker_id"]
respectively?
Did you try to call the second image separately to see what is coming out of the
database?
Hope this helps
Dobromir Velev
Web Developer
http://www.websitepulse.com
-Original Message-
From: Christopher Martin <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
Date: Thursday, December 20, 2001 19:08
Subject: [PHP-DB] Multiple Image Display Problem
>I am having problems displaying two different images on the same page.
>(stored as binary files in two different db tables)
>
>Instead of getting two different images, I am getting the first image twice,
>although the scripts are going to two different database tables to get the
>two images.
>
>Is there a way to get both images to display on this page?
>
>Any help would be much appreciated. Thank you.
>
>The main code is:
>if (count($result3) > 0) {
>while ($result_array3 = mysql_fetch_array($result3, MYSQL_ASSOC)) {
>if ($result_array3["speaker_id"] !== "") {
>echo "$result_array["speaker_id"] . "\">";
>}
>}
>}
>if (count($result4) > 0) {
>while ($result_array4 = mysql_fetch_array($result4, MYSQL_ASSOC)) {
>if ($result_array4["speaker_id"] !== "") {
>echo "$result_array["speaker_id"] . "\">";
>}
>}
>}
>?>
>-
>small_image.php:
>
>include("include.inc");
>$sql = "SELECT bin_data, filetype, filename, filesize FROM smphoto_bin_data
>WHERE speaker_id=$speaker_id";
>// . $HTTP_SESSION_VARS["speaker_id"];
>
>$result = @mysql_query($sql);
>$data = @mysql_result($result, 0, "bin_data");
>$name = @mysql_result($result, 0, "filename");
>$size = @mysql_result($result, 0, "filesize");
>$type = @mysql_result($result, 0, "filetype");
>
>header("Content-type: $type");
>header("Content-length: $size");
>header("Content-Disposition: attachment; filename=$name");
>header("Content-Description: PHP Generated Data");
>echo $data;
>?>
>
>large_image.php:
>
>include("include.inc");
>$sql = "SELECT bin_data, filetype, filename, filesize FROM lgphoto_bin_data
>WHERE speaker_id=$speaker_id";
>
>$result = @mysql_query($sql);
>$data = @mysql_result($result, 0, "bin_data");
>$name = @mysql_result($result, 0, "filename");
>$size = @mysql_result($result, 0, "filesize");
>$type = @mysql_result($result, 0, "filetype");
>
>header("Content-type: $type");
>header("Content-length: $size");
>header("Content-Disposition: attachment; filename=$name");
>header("Content-Description: PHP Generated Data");
>echo $data;
>?>
>
>
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