mysql_fetch_array(): supplied argument is not a valid MySQL result

[sherdog]

Im having a bit of trouble, I redefined my query and im getting this error,

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\inetpub\wwwroot\ecg\SSL\orders_review.php

here is my query:

$status_query3 = "select *
from
customer
join hosting_info
on (customer.status = hosting_info.status)
left join billing
on (customer.status = billing.status)
right join billing_status
on (billing.status = billing_status.status)
join orders
on (customer.status = orders.status)
where customer.status=1";

I ran a if(mysql_query(status_query3){
echo "succesful query";
}else{
echo mysql_error();
}

and i get a successful query
so i know it isnt the query

here is my loop

echo "<table width='100%' cellpadding='3' cellspacing='0' border='0'>";
echo "  <tr>";
echo "    <td colspan='9'>Pending Accounts = ".$totalorder."</td>";
echo "  </tr>";
echo "  <tr>";
echo "    <th align='left'><strong>Customer Name</strong></th>";
echo "    <th align='left'>Billing Email</th>";
echo "    <th align='left'>Total Price</th>";
echo "    <th align='left'>Date yyyy-dd-mm</th>";
echo "    <th align='left'>Package</th>";
echo "    <th align='left'>Payment Term</th>";
echo "    <th align='left'>Billing Type</th>";
echo "    <th align='left'>Domain name</th>";
echo "    <th align='left'>Details</th>";
echo "  </tr>";
$color1  = "#DDF4FF";
$color2 = "#FFFFFF";
$row_count = 0; 

while( $myrow = mysql_fetch_array($result) ){
	$row_color = ($row_count % 2) ? $color1 : $color2; 
	echo "  <tr>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['fname']." ".$myrow['lname']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['b_email']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['price']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['tdate']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['package']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['billing_term']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['payment_method']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap>".$myrow['domain']."</td>";
	echo "     <td bgcolor=".$row_color." nowrap><a href='reviewdetails.php?cid=".$myrow['cid']."'><img src='images/lastpost.gif' border='1'></a></td>";
	echo "   </tr>";
	$row_count++; 
}

echo "</table>";

so I'm thinking that maybe because my query contains multiple columns with the same name, but i thought that is what defined the foreign key.

please help
=)

[ctimmer]

Where are you setting $result?

[sherdog]

whoops,

here is the result with the query error checker and it is before the while loop and after the $status_query3

$result = mysql_query($status_query3);

if (mysql_query($status_query3) ){
echo "Successful query";
}
else{
echo mysql_error();
}

[sherdog]

I also tried a different query instead of using joins, i just made a equi-join
$status_query2 = "SELECT
c.fname, c.lname, c.b_email, c.cid, c.status, o.status, o.price, h.status,
b.status, h.billing_term, h.domain, b.payment_method o.tdate, o.status FROM

customer c, orders o, hosting_info h, billing b
WHERE c.status = 1 AND o.status = 1 AND b.status = 1 AND h.status = 1";

And i am still getting this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\inetpub\wwwroot\ecg\SSL\orders_review.php
why is it invalid when i test the query and it is a successful query?

[ctimmer]

I don't know if mySQL permits multiple active result sets (you create a second one with the 'if (mysql_query($status_query3) )' statement and discard it).

Try placing the '$result = ...' statement after the 'if (mysqlquery(...' statement.

[sherdog]

nope didn't work either... hmmm

[Threehearts]

he doesnt accept your sql syntax...
try WHERE some_id = 'value'
... it never worked with me when i left out the ' '

WHERE c.status = '1' AND o.status = '1' AND b.status = '1' AND h.status = '1'";

maybe thats the problem, try it.
greetings, jakob

[sherdog]

well what i did was made a error checking statement and i found it was my query after all =\

i rewrote it to this $status_query2 = "SELECT * FROM customer, orders, billing, billing_status, hosting_info
WHERE customer.status = '1' AND
orders.cid = customer.cid
AND billing.cid = customer.cid
AND billing_status.cid = customer.cid
AND hosting_info.cid = customer.cid";

and it works now thanks for the help