Try rewriting your query. It looks to me that you probably have an
error in your SQL statement and it only seems fine because you appended
the @ to the beginning of the mysql_query() function.
I would suggest using a join statement.
If(!$result = mysql_query("SELECT orders.orderid, orders.amount,
orders.date, customer.name FROM orders RIGHT JOIN customers on
orders.customerid = customers.customerid WHERE customers.name = 'Don
Hansen'")) {
echo(mysql_error()); //I would take this out after you are done
debuging
} else {
while ($row = mysql_fetch_array($result)) {
$zorderid = $row['orders.orderid'];
$zamt = $row['orders.amount'];
echo(str_pad($zorderid,4,' ').str_pad($zamt,6, ' ').'<br>');
}
}
-----Original Message-----
From: ddhull@ku.edu [mailto:ddhull@ku.edu]
Sent: Tuesday, October 28, 2003 9:20 AM
To: php-db@lists.php.net
Subject: show data from 2 tables
I am attempting to pull data from 2 tables. My SELECT statement seems
ok. But the next line: while ... etc. is giving me an error of
"Supplied argument is not a valid MySQL result resource.". But this is
the way I usually show my data and it works fine when only 1 table is
involved. Is there a different way of showing your data when pulling
data from more than one table?
$result = @mysql_query('SELECT orders.orderid, orders.amount,
orders.date
from customers, orders WHERE customers.name = "Don Hansen" and
customers.customerid = orders.customerid');
while ( $row = mysql_fetch_array($result) )
{
$zorderid = $row['orders.orderid'];
$zamt = $row['orders.amount'];
echo (str_pad($zorderid,4,' ') . str_pad($zamt,6, ' ') .
'<br>');
}
Thanks,
Doug
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