Buggy SQL/PHP?

[Anon]

I need a second pair of eyes to look over what I am doing to see if I fubar'd something, cause I sure cant find it.

<?php
$link = mysql_connect("localhost", "login", "passwd") or die("Could not connect");
mysql_select_db("database_name", $link) or die("Could not select database");

    $query_table = mysql_query("SELECT Email, Password, AcctType, IPAddress FROM customer",
            $link);

    print "<table>";
    print "<tr align=left>";
    print "<th>E-Mail</th><th>Password</th><th>Acct Type</th><th>IP Address</th></tr>";

    while ($row = mysql_fetch_object($query_table))
    {
            print "<tr>";
            print "<td>" . $row->Email . "</td>";
            print "<td>" . $row->Password . "</td>";
            print "<td>" . $row->AcctType . "</td>";
            print "<td>" . $row->IPAddress . "</td>";
            print "</tr>";
    }
    print "</table>";

?>

Error given:

Warning: Supplied argument is not a valid MySQL result resource in imm.php on line 12

What did I do wrong or typo on?

-- M

[[deleted]]

Everything looks good as far as I can see. I would check that query and make sure it is valid. If it's not it will return FALSE and then $query_table will contain FALSE rather than being a result resource. Maybe make this change:

$query_table = mysql_query("SELECT Email, Password, AcctType, IPAddress FROM customer",
$link) or die("Invalid query.");

Matt Wade
codewalkers.com

[zalath]

A good habit is to check the queries by putting print mysql_errno().": ".mysql_error(); after the query. It will show both the error code and the mySQL error message.

-L-

[[deleted]]

indeed.

remove the 'or die', and put something like this instead:

if (!$result = mysql_query('query'))
{
// print error
echo mysql_error();
}
else
{
// Process results
}