I am trying to select data from a MySQL table, but I get one of the following error messages:
mysql_fetch_array() expects parameter 1 to be resource,
As scompt.com explained, the query might fail. Use this code the get the error of the query or the correct result:
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("
SELECT * FROM Users
WHERE UserName LIKE '".mysql_real_escape_string($username)."'
");
if($result)
{
while($row = mysql_fetch_array($result))
{
echo $row['FirstName'];
}
} else {
echo 'Invalid query: ' . mysql_error() . "\n";
echo 'Whole query: ' . $query;
}
See the documentation for mysql_query() for further information.
The actual error was the single quotes so that the variable $username was not parsed. But you should really use mysql_real_escape_string($username) to avoid SQL injections.