Warning: mysql_result(): supplied argument is not a valid MySQL result resource in (…) on line 4

Question

Hi guys just curious to solve this annoying problem. Heres my snippet.

I\'ve checked some other questions similar to my error but so far I cant get it solved.

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Answer 1

This could happen, when mysql_query returns false, if it fails for some reason. So you should split this into multiple statements and check the return values

$sql = "SELECT COUNT(user_id) FROM users WHERE username =  $username";
$result = mysql_query($sql);
if ($result === false) {
    // error handling
    return false;
}

return (mysql_result($result, 0) == 1) ? true : false;