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;

Answer 2

You should split your code in some more lines to handle those errors or special cases. mysql_query will return zero to n rows or an error if it occurs. The returned resource will therefore only be true on non-error queries. This can be used to handle such situations like follows.

At first build and execute query, next process the resource.

$query="SELECT COUNT(user_id) FROM users WHERE username = ".$username;
$result = mysql_query($query);  

u may use the following to determine what is going on in case of an error:

if(!$result) die("SELECT failed: ".mysql_error());

or these idea to handle the problem

if (!$result=mysql_query($query)) {
        return false; // or similar operation
    }

    if (mysql_num_rows($result)!=1){
        return false;
    }else{
        return true;
    }