Error in On-line [Duplicate]

Do this thing:

$time = time();
$online = $time - (20*60);
$result = mysql_query("SELECT username FROM users WHERE users_come >= $online");
while( $row = mysql_fetch_array( $result ) ){
echo "<br>{$row['username']}"; }

She displays users on Oolain
But I knock out such a mistake:

warning: mysql_fetch_array () EXPECTS PARAMETER 1 TO BE RESOURCE, BOOLEAN GIVEN IN C: XAMPPHTDOCSSTYLEONLINE.PHP ON LINE 20

What does it mean and how to fix tell me please!


Answer 1, Authority 100%

Most likely you could not connect to the database. Or the request is not true. For example, there is no field or it is called differently.

Make after

$result = mysql_query("SELECT username FROM users WHERE users_come >= $online");
var_dump(mysql_error());