$ stmt-＆gt;错误throws尝试获取非对象的属性

Question

我有以下查询：

// pagination
if ($stmt = $mysqli->prepare('SELECT COUNT(id) FROM ' . $table)) {
    $stmt->execute();

    $stmt->bind_result($totalitems);

    $stmt->fetch();

    if (!isset($page)) {
        $page = (int)$page_number <= 0 ? 1 : (int)$page_number;
    }

    $limit = 4;

    if ($page > ceil($totalitems / $limit)) {
        $page = ceil($totalitems / $limit);
    }

    $start = ($page - 1) * $limit;

    $stmt->close();

    if ($stmt2 = $mysqli->prepare(' SELECT t1.id, t2.*
                                    FROM ' . $table . ' t1
                                    INNER JOIN ' . $table2 . ' t2 ON t2.id = t1.id
                                    ' . $optional_cond . $optional_cond2 . '
                                    LIMIT ?, ?')) {
        $stmt2->bind_param('ii', $start, $limit);
        $stmt2->execute();

        $stmt2->bind_result($col1, $col2, etc...);

        while ($row = $stmt2->fetch()) {
            echo $col1 . '<br>';
        }
    } else echo "Statement failed: " . $mysqli->error . "<br>";
}

它无效，所以我添加了else：

else echo "Statement failed: " . $stmt2->error . "<br>";

但它发出以下通知：

Notice: Trying to get property of non-object

我无法查看我的查询有什么问题，发生了什么？

FIX（由@Michael Berkowski撰写 - 评论中）：

else echo "Statement failed: " . $mysqli->error . "<br>";