Warning when calling mysqli_error() [duplicate]

Question

Possible Duplicate:
PHP warning help?

I'm trying to join three tables from a database in-order to display a users selected categories but I get the following error.

Warning: mysqli_error() expects exactly 1 parameter, 0 given in

I think I'm doing something wrong when I query my database.

Here is the code below.

// Query member data from the database and ready it for display
$mysqli = new mysqli("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT users.*, categories.*, users_categories.* FROM users_categories 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");

if (!$dbc) {
    // There was an error...do something about it here...
    print mysqli_error();
}   

//Users entered category loop
while ($row = mysqli_fetch_assoc($dbc)) {
        if (! empty($row['category'])) {
                echo '<div class="skill-info">';
                echo '<p>' , htmlspecialchars($row['category']) , '</p>';
            }

    }

Here is my MySQL tables structure.

CREATE TABLE users (
user_id INT UNSIGNED NOT NULL AUTO_INCREMENT,
user_name VARCHAR(255) NOT NULL,
PRIMARY KEY (user_id)
);

CREATE TABLE categories ( 
id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
parent_id INT UNSIGNED NOT NULL DEFAULT 0, 
category VARCHAR(255) NOT NULL, 
url VARCHAR(255) NOT NULL, 
PRIMARY KEY (id), 
INDEX parent (parent_id)
);

CREATE TABLE users_categories (
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
user_id INT UNSIGNED NOT NULL, 
category_id INT UNSIGNED NOT NULL, 
PRIMARY KEY (id)
);

Now I get the following error

Not unique table/alias: 'users_categories'

How do I fix this?

Thanks everyone for the help. Is there a way i can reward every one for there help here on stackoverflow instead of one person?

Answer 1

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.