Warning when calling mysqli_error() [duplicate]

Answer 1

mysqli_error requires a link, try:

print mysqli_error($mysqli);

Answer 2

You should call mysqli_error with parameter:

if (!$dbc) {
    print mysqli_error($mysqli);
}   

EDITED:

You join two table with the same names. I think in FROM you made typo. Shouldn't it be FROM users? Table users appears in column list but you do not have it anywhere in FROM or JOINs.

SELECT users.*, categories.*, users_categories.* 
FROM users
INNER JOIN users_categories ON users_categories.user_id = users.user_id 
JOIN categories ON users_categories.user_id = users.user_id 
WHERE users_categories.user_id=3

Answer 3

You forgot to pass a required argument to mysqli_error(). Change

print mysqli_error();

to

print mysqli_error($mysqli);

Once you make the correction, you'll be able to see the error message and debug further.

FYI: It looks like you're oddly mixing the OO and procedural style usage of mysqli. You should stick to either one or the other.

Answer 4

There are lots of comments on how to fix the mysqli_error line, but there are two problems with the SQL itself:

1. You are joining the users_categories table twice and not joining the users table at all.
2. The second ON is identical to the first, so it references the wrong columns in the wrong tables.

This should work better:

SELECT users.*, categories.*, users_categories.*
FROM users_categories
INNER JOIN users ON users_categories.user_id = users.user_id 
INNER JOIN categories ON users_categories.category_id = categories.id
WHERE users_categories.user_id=3

Answer 5

The query is the problem. Test the query first in PhpMyAdmin or whatever mySQL client you have at your disposal, and you'll see it doesn't work.