Warning: mysql_result() expects parameter 1 to be resource, boolean given

Question

i know that we already had a lot of question about this error, but i can not fix my code, so anyone here please help me fix this problem. my code like this

function login($username, $password) {
    $user_id = user_id_from_username($username);

    $username = sanitize ($username);
    $password = md5 ($password);

    return (mysql_result(mysql_query("SELECT COUNT (`user_id`) FROM `users` 
        WHERE `username` = '$username' AND `password` = '$password'"),0)==1) ? $user_id : false;

i am looking forward to hear from you guys. i am a new member on this so please keep your answer as simple as possible. thank you.

Echo your query instead of executing it, copy it, paste it into phpmyadmin or something similar and see if the query executes. — Epodax, Nov 25 '15 at 07:53
http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php?rq=1 — Sebastian Brosch, Nov 25 '15 at 08:00
Why the hell you first find the user_id by username and after that you make a second query to verify the password instead of making it at once? And between those 2 actions you sanitize the username (so it may become even different between the 2 queries..) — Svetoslav, Nov 25 '15 at 08:03
i am just study from scratch, and i follow this one on youtube, i dont know why in this video they could do it but i can not. — Hoàng Trung Hiếu, Nov 25 '15 at 08:33

score 1 · Answer 1 · answered Apr 05 '21 at 10:50

1

I think that this problem is occured because : You write mysql every where. Instead Of that write it mysqli every where. If till now also the problem is not resolved. Let me know I will tell another method to fix this issue.

answered Apr 05 '21 at 10:50

Tejaswa

11
10

score 0 · Answer 2 · answered Nov 25 '15 at 08:13

0

Here is a little remake of you method.

! Important - MYSQL functions should be replaced with MYSQLi as MYSQL driver is deprecated after php 5.5 and removed in the new PHP7...

function login($username, $password) {
    $username = sanitize ($username);
    $password = md5 ($password);

    $sql = "SELECT `user_id` FROM `users` WHERE `username` = '{$username}' AND `password` = '{$password}'";
    $result = mysql_query($sql);
    if( $row = mysql_fetch_assoc($result)) {
        return $row['user_id'];
    }
    return FALSE;
}

answered Nov 25 '15 at 08:13

Svetoslav

4,686
2
28
43

i follow this one on the video from youtube, they could run it while i can not. – Hoàng Trung Hiếu Nov 25 '15 at 08:37
Whats not working ? What do you want from this function ? – Svetoslav Nov 25 '15 at 08:43
the thing i want is that, if the username and the password is true so i can login. – Hoàng Trung Hiếu Nov 25 '15 at 08:45
`$login = login($username, $password); if ($login === false){ $errors[] = 'That username/password combination is incorrect'; }else{ echo 'ok!';` – Hoàng Trung Hiếu Nov 25 '15 at 08:46
then can you verify that the user name and the md5 of the password are correct ? with var_dump($sql) check what is the generated query and test it at your phpmyadmin ... – Svetoslav Nov 25 '15 at 08:50
my username and password are correct i am sure about it because i made username and password are the same – Hoàng Trung Hiếu Nov 25 '15 at 08:58

score 0 · Answer 3 · answered Nov 25 '15 at 08:14

0

mysql_query will return false on error, and a resource on success. You need to check the return value of mysql_query before calling mysql_result.

$resource = mysql_query("SELECT COUNT (`user_id`) FROM `users` WHERE `username` = '$username' AND `password` = '$password'") or die(mysql_error());
if($resource !== false) 
{
    return (mysql_result($resource,0)==1) ? $user_id : false;
}   
return false

I would also escape the input if you have not done so already using mysql_real_escape_string()

answered Nov 25 '15 at 08:14

Daryl B

525
5
16

it does not work for me, still error, i did follow this one on a video, they can run it while i can not. – Hoàng Trung Hiếu Nov 25 '15 at 08:42
I would suspect you are getting a database error if you are still getting that warning. Perhaps echo out the SQL and run it manually and solve the SQL error. Once this is working properly, then you can rule that out. – Daryl B Nov 25 '15 at 22:45

score 0 · Answer 4 · answered Dec 03 '17 at 09:48

Thats error means that all in mysql_result( HERE ) is returns false not a resourse!

You need to var_dump, what final result you get in your query then try execute in via PMA of via console from mysql>
You will see that nothing is returned or error in query.

then fix your query and/or code and be cool!

if too hard, post in comments your resul query, i'll fix it !

score -1 · Answer 5 · answered Nov 25 '15 at 07:54

-1

Remove the last ` on your query next to false.

answered Nov 25 '15 at 07:54

William Madede

727
4
8

where is the symbol ` that you mean? can you pls fix it and print the right code in here for me? – Hoàng Trung Hiếu Nov 25 '15 at 08:01
return (mysql_result(mysql_query("SELECT COUNT (`user_id`) AS count FROM `users` WHERE `username` = '$username' AND `password` = '$password'"),0)==1) ? $user_id : false; – William Madede Nov 25 '15 at 08:02
thank you for your help but it still not work for me :(( – Hoàng Trung Hiếu Nov 25 '15 at 08:05

Warning: mysql_result() expects parameter 1 to be resource, boolean given

5 Answers5