Ошибка Warning: mysqli_error() expects exactly 1 parameter, 0 given в запросе

Помогите исправить ошибку в запросе:

Warning: mysqli_error() expects exactly 1 parameter, 0 given in D:\OpenServer\OSPanel\domains\train2\second.php on line 29
ERROR QUERRY

index.php

<form action="second.php" method="POST">
    <fieldset>
    <label for="first_name">first name</label>
    <input type="text" name="first_name"> <br>
    <label for="first_name">last_name</label>
    <input type="text" name="last_name"> <br>
    <label for="first_name">bio</label>
    <input type="text" name="bio"> <br>
    <label for="first_name">email</label>
    <input type="text" name="email"> <br>
    <label for="first_name">facebook_url</label>
    <input type="text" name="facebook_url"> <br>
    <label for="first_name">twitter_url</label>
    <input type="text" name="twitter_url"> <br>
    </fieldset>
    <fieldset>
        <input type="submit" name="ok">
    </fieldset>
</form>

second.php

<?php 
require "db_connection.php";
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$bio = $_POST['bio'];
$email = $_POST['email'];
$facebook_url = $_POST['facebook_url'];
$twitter_url = $_POST['twitter_url'];
$show_tb = mysqli_query($connection, "SHOW TABLES");
$sel_tb = mysqli_query($connection, "USE TABLES my");
if ($sel_tb) {
    mysqli_error();
}else{
    echo "<br> okkkkkkkkkkk <br>";
}
$qq = mysqli_query($connection,"SELECT * FROM my");
if (!$qq) {
    echo "noooooo" . mysqli_query();
}else{
    echo "select good";
}
$query = mysqli_query($connection, "INSERT INTO my(first_name, last_name, bio, email, facebook_url, twitter_url) VALUES ($first_name, $last_name, $bio, $email, $facebook_url, $twitter_url)");
if (!$query) {
    die("ERROR QUERRY" . mysqli_error());
}
while ($row = mysqli_fetch_row($query)) {
echo "You'e use  {$row[0]}";
}
mysqli_close($connection);
 ?>

Подключение к БД есть. Только не могу сделать запрос в БД посредством ввода информации в инпут.