need help

[chinwah]

I wrote this code:

$db = new DB_Example;

$query = "SELECT EXP_COM_NAME, EXP_HIRED_DATE from experience where CAN_ID = '$user->id'";
$result = $db->query($query);
:83 while ($row = mysql_fetch_array ($result)){
$query2 = sprintf ("select PRO_TITLE from project where CAN_ID='$user->id' and EXP_COM_NAME= '%s'", $row["EXP_COM_NAME"]);
$result2 = $db->query($query2);
while ($row2 = mysql_fetch_array ($result2)){
............
............
}
}
So here is the error
Warning: 6 is not a valid MySQL result
resource in /file on line 83

I use PHPLIB as the database abstraction layer.

what is wrong?

thank you

[jbauer23]

Try running that query from the mysql prompt and see if it works. Usually that error occurs with an invalid query.