help with javascript (moved from javascript forum)

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help with javascript

 
0
  #1
Jul 5th, 2005
http://ww.bestblanks.co.uk/media.php

hi

im not sure what im doing wrong with this:

have a look at the above url:

each product name is suppose to open a small window which will go on to display a description of the product: however when you click on title there is an error and nothing loads

can anybody suggest what im doing wrong

thanks

simon

this is the code for media.php:

<?php include "vsadmin/db_conn_open.php"; ?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>bestblanks.co.uk</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<link href="bluetipdvd/bluetipdvd.css" rel="stylesheet" type="text/css">
<script language="javascript">
function longdescription(ID) {
url = 'proddetail.php?ID='+ID;
window.open(url,'welcome','width=400,height=200');
}
</script>
<SCRIPT SRC="bluetipdvd/language-en.js"></SCRIPT>
<SCRIPT SRC="bluetipdvd/nopcart.js"></SCRIPT>
</script>
</head>
<body background="bluetipdvd/images/background.jpg" leftmargin="0" topmargin="0" marginwidth="0" marginheight="0">
<table width="509" height="100%" border="0" align="center" cellpadding="1" cellspacing="1">
<tr class="rightborder">
<td width="505" valign="top"> <table width="100%" height="100" border="0" align="center" bgcolor="#FFFFFF" valign="top">
<?php
$cat = $_GET['cat'];
if(empty($cat)) {
$query = "SELECT * FROM products ORDER BY 'price'";
} else {
$query = "SELECT * FROM products WHERE cat = '$cat' ORDER BY 'price' ";
}
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
$num=mysql_num_rows($result);

mysql_close();


$i=0;
while ($i < $num) {

$ID=mysql_result($result,$i,"ID");
$name=mysql_result($result,$i,"name");
$price=mysql_result($result,$i,"price");
$photo=mysql_result($result,$i,"photo");
$longdescription=mysql_result($result,$i,"longdescription");
$qty=mysql_result($result,$i,"qty");
?>
<tr>
<td width="31%" rowspan="2" valign="top" class="name"><div align="left"></div>
[<? echo "$ID" ?>] <br> <img src="<? echo "$photo" ?>" alt="<? echo "$name" ?>"></td>
<td width="69%" valign="top" class="name"><p><span class="name"><a href="javascript:longdescription(<?= $ID ?>)">
<?=$name?>
</a> </p>
<p><span class="name"> Our price £<? echo "$price" ?></span></p>
<p>&nbsp;</p></td>
</tr>
<tr>
<td valign="top"> <p>
<form name=order>
<input type=hidden name="qty" onChange='this.value=CKquantity(this.value)' value="1">
<input type=hidden name="PRICE" value="<?php echo "$price" ?>">
<input type=hidden name="NAME" value="<?php echo "$name" ?>">
<input type=hidden name="ID_NUM" value="<?php echo "$ID" ?>">
<input type=hidden name="SHIPPING" value="0.00">
<input type=image src="../images/addcart.gif" border="0" onClick='AddToCart(this.form)'>
</FORM></form>
</td>
</tr>
<?
$i++;
}
?>
</table></td>
</table>
</body>
</html>

and this is the code for the window that opens:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<?php include "vsadmin/db_conn_open.php";
$ID = $_GET['ID'];

$query = "SELECT * FROM products";
}
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
$num=mysql_num_rows($result);

mysql_close();


$name=mysql_result($result,$i,"name");
$price=mysql_result($result,$i,"price");
$longdescription=mysql_result($result,$i,"longdescription");
?>

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



<body>
<table width="100%" height="91" border="1" cellpadding="1" cellspacing="1">
<tr>
<td><strong><font color="#0000FF" size="2" face="Arial, Helvetica, sans-serif">[</font><font color="#0000FF" size="2" face="Arial, Helvetica, sans-serif"><? echo "$ID"?>
] </font></strong></tr>
<tr><td><? echo "$name"?></td></tr><tr><td><? echo "$longdescription"?></td></tr>
<td height="18"></tr>
</table>

</body>
</html>
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Join Date: Dec 2004
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tgreer tgreer is offline Offline
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Re: help with javascript

 
0
  #2
Jul 5th, 2005
You seem to be asking someone to troubleshoot your PHP code. So, I've moved this thread to the PHP forum.

If you're asking a JavaScript question, post only the JavaScript code that appears to contain the problem.
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Re: help with javascript (moved from javascript forum)

 
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  #3
Jul 5th, 2005
Hello tgreer,
i am not agree with u
on his code there is also PHP.
so why move his post?
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tgreer tgreer is offline Offline
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Re: help with javascript (moved from javascript forum)

 
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  #4
Jul 5th, 2005
Because I saw no JavaScript in his code. I saw PHP code that produces JavaScript. So the first step in troubleshooting the problem is to make sure all of the PHP is valid. If it is, then we can move on to working through the JavaScript.

If a user truly wants help with JavaScript, they should post the relevant JavaScript. With server-side-generated code, the best way to see what is actually produced is to "view source" and copy the relevant code snippets to the forum. Which the original poster can still do! He hasn't been "banned" from the JavaScript forum by any means. I just made a judgment call that he would get better help here.

While we're talking about clarity, in a technical forum I would recommend proper spelling and punctuation.
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