将参数动态绑定到mysqli语句

我正试图构建一个动态准备语句，但我一直停留在bind_param部分。我试着阅读其他关于call_user_func_array的答案，但我不知道如何在这里进行调整：

//connecting
$connection = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME);
//info submitted through a form
$values_columns = array(
                        "column_one" => "value_one",    //value_one will be replaced by $_POST['value_one']
                        "column_two" => "value_two",    //value_two will be replaced by $_POST['value_two']
                        "column_three" => "value_three" //value_three will be replaced by $_POST['value_three']
                       );
$value_type = implode('', array('s', 's', 's'));
//preparing and binding my query dinamically
function prepare_bind($table_name, $values_columns, $value_type) {
    global $connection;
    $columns_string = "";
    $question_marks = "";
    $flag = 0;
    $count = count($values_columns);
    foreach ($values_columns as $column => $value) {
        $flag++;
        // building the prepare
        if ($flag == $count) {
            $columns_string .= $column; 
            $question_marks .= "?";
        } else {
            $columns_string .= $column . ", ";  
            $question_marks .= "?, ";
        }
    }
    $sql = $connection->prepare('INSERT INTO ' . $table_name . ' (' . $columns_string . ') VALUES (' . $question_marks . ')');
    $sql->bind_param($value_type, /*I am stuck here while trying to add the values*/);
}

有什么办法摆脱这种局面吗。这是我第一次使用这种方法，我不知道我是否做对了。非常感谢。