PHP problem

meo_beo · 05-16-2008, 03:18 AM

Hi guys, I'm trying to display all the records in a table, say the 'order' table, of my MySql database on a webpage using PHP. Here is the source code:

<?php
include('conn.php');
$sql = "select * from order";
$result = mysql_query($sql,$link);
?>

<body>
<center><h4>Details of each order</h4></center>
<div align="center">
<table border="1">
<tr>
<td>Order ID</td>
<td>Order Date</td>
<td>Customer ID</td>
</tr>
<?php
$total_row = mysql_num_rows($result);
if ($total_row > 0){
$i = 0;
while ($row = mysql_fetch_array($result)){
$i +=1;
?>
<tr>
<td><?= $row[0]?></td>
<td><?= $row[1]?></td>
<td><?= $row[2]?></td>
</tr>
<?php
}

}
else{
echo "Error!";
}
?>
</table>
The problem is, it always produces such an error:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in
It confuses me because if I change the sql statement to "select * from order_detail", which means I only change the table name, then it works just fine. So I believe the problem is with the "order" table itself. Any suggestions?

datopdog · 05-16-2008, 03:26 AM

It means your query failed, you need to test $result first before trying to use it.

Wim Sturkenboom · 05-16-2008, 04:56 AM

To add to that, your problem can already be in conn.php; was mysql_connect successfull? Did you use mysql_select_db to select a database and was it successfull?

Further mysql_error can be of assistance to determine the error. See http://www.php.net/mysql_error

Always test the return values of functions !!

michaelk · 05-16-2008, 07:06 AM

Your problem is due to order being a MySQL reserved or keyword. Try using quotes like:

$sql = "select * from `order`";
or
$sql = "select * from \"order\"";

Or do not use keywords for table names.

http://dev.mysql.com/doc/refman/5.0/...ved-words.html

BTW use code tags in the future.

meo_beo · 05-18-2008, 06:31 AM

Thank you, michaelk, I think that's what the problem is.