I'm trying to display two images on one php page from 2 different database tables. It is displaying the first image twice. Has anyone encountered this before? Any work-arounds?
Thanks in advance.
The code:
display.php:
<tr valign="top">
<td><IMG SRC="small_image.php?speaker_id=<? echo $speaker_id ?>"></td>
</tr>
</table>
<table border="1" width="300">
<tr valign="top">
<td width="100"><p>Large Image:</p></td>
</tr>
<tr valign="top">
<td><IMG SRC="large_image.php?speaker_id=<? echo $speaker_id ?>"></td>
</tr>
small_image.php:
<?
include("include.inc");
$sql = "SELECT bin_data, filetype, filename, filesize FROM smphoto_bin_data WHERE speaker_id=$speaker_id";
$result = @($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");
$sql2 = "SELECT bin_data, filetype, filename, filesize FROM lgphoto_bin_data WHERE speaker_id=$speaker_id";
$result2 = @($sql2);
$data = @mysql_result($result2, 0, "bin_data");
$name = @mysql_result($result2, 0, "filename");
$size = @mysql_result($result2, 0, "filesize");
$type = @mysql_result($result2, 0, "filetype");
header("Content-type: $type");
header("Content-length: $size");
header("Content-Disposition: attachment; filename=$name");
header("Content-Description: PHP Generated Data");
echo $data;
?>
