No luck with popup slideshow

DirectMe · Apr 16, 2003

I am trying to create a popup slideshow. Everything seems to check out but no data is displayed in the popup window except for the info passed in the querystring. (ie $listingnumber in title is correct). Can anyone show me where I am going wrong? I am Stumped.

The formatting shows but no images or text. Also $id remains blank. AAAAAAAAAAAAAarggh!! What am I doing wrong?? [sadeyes]

<html>
<head>
<title>Listing <? echo $listingnumber ?>Slideshow</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<style>

</style>
</head>
<body>
<table width="95%" border="0" cellspacing="1">
<tr align="center" valign="top">
<td>
<table width="80%" border="0" cellspacing="1">
<tr>
<td><div align="center"><IMG SRC="<?php echo $image;?>" border="1" width="350" height="300" alt="sorry image not available"></div>
</td>
</tr>
<tr>
<td><div align="center"><?php echo $temp[caption][$id];?><?php echo $id; ?></div>
</td>
</tr>
<tr>
<td><div align="center">
<a href="javascript:history.back()"><<<</a>
| <a href="view_images.php?listingnumber=<?php echo $listingnumber; ?>&id=<?php echo $id; ?>">>>></a></div>
</td>
</tr>
</table>
</td>
</tr>
</table>

<?php
$dbh=mysql_connect ("HOSTNAME", "USERNAME", "PASSWORD&quot

or die ('I cannot connect to the database because :'. mysql_error());
mysql_select_db ("DATABASE", $dbh);
$result = @ mysql_query("SELECT caption, imgname FROM directho_.listing_images WHERE listingnumber = $listingnumber&quot

;
if ($result)
{
$rows=mysql_num_rows($result);
$temp=array();

for ($i=0; $i<$rows; $i++)
$temp[]= mysql_fetch_array($result);

$uploaddir = "/listing_images/" . $listingnumber . "/";

if ($id=="&quot

{
$id=0;
}elseif ($id==$rows){
$id=0;
}else{
$id=($id+1);
}
$image = $uploaddir.$temp

[$id];
}
else
{ print "Sorry there are no pictures on this listing: $listingnumber<br><br>";
}
MYSQL_CLOSE();
?>
</body>
</html>

sleipnir214 · Apr 16, 2003

I don't see where you actually output anything you fetch from the database.

Want the best answers? Ask the best questions:

http://www.catb.org/~esr/faqs/smart-questions.html

TANSTAAFL!

DirectMe · Apr 16, 2003

I am trying to output by echo in the table:

IMG SRC="<?php echo $image;?>"
and
<?php echo $temp[caption][$id];?><?php echo $id; ?>

I know its pretty muddy code, but, I a very new to php.

sleipnir214 · Apr 16, 2003

Start at the top of your script and work your way through it.

Your script outputs some HTML. In there is a command to print $image.

Then later in your script, you actually assign $image a series value. But you do that only after you have already printed your HTML

Want the best answers? Ask the best questions:

http://www.catb.org/~esr/faqs/smart-questions.html

TANSTAAFL!

KempCGDR · Apr 16, 2003

Basically, you can't print something before you fetch it from the database.

That'd mean your code could see into the future and Micro$oft would have already copied your code, patented it and sued you.

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No luck with popup slideshow

DirectMe

IS-IT--Management

sleipnir214

Programmer

DirectMe

IS-IT--Management

sleipnir214

Programmer

KempCGDR

Programmer

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