php - fatal error : Call to a member function fetch_assoc() on a non-object

Question

这个问题已经有答案了:

What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such (1 个回答)

已关闭 4 年前。

我正在尝试执行一些查询来获取有关某些图像的信息页面。我写了一个函数

function get_recent_highs($view_deleted_images=false)
{
    $lower = $this->database->conn->real_escape_string($this->page_size * ($this->page_number - 1));
    $query = "SELECT image_id, date_uploaded FROM `images` ORDER BY ((SELECT SUM( image_id=`images`.image_id ) FROM `image_votes` AS score) / (SELECT DATEDIFF( NOW( ) , date_uploaded ) AS diff)) DESC LIMIT " . $this->page_size . " OFFSET $lower"; //move to database class
    $result = $this->database->query($query);
    $page = array();
    while($row = $result->fetch_assoc())
    {
        try
        {
            array_push($page, new Image($row['image_id'], $view_deleted_images));
        }
        catch(ImageNotFoundException $e)
        {
            throw $e;
        }
    }
    return $page;
}

根据这些图像的受欢迎程度选择这些图像的页面。我写了一个Database处理与数据库和 Image 交互的类保存有关图像信息的类。当我尝试运行此程序时，出现错误。

Fatal error: Call to a member function fetch_assoc() on a non-object

$result是一个 mysqli 结果集，所以我很困惑为什么这不起作用。

Answer 1

那是因为您的查询中有错误。 MySQli->query()出错时将返回 false。将其更改为类似以下内容::

$result = $this->database->query($query);
if (!$result) {
    throw new Exception("Database Error [{$this->database->errno}] {$this->database->error}");
}

如果出现错误，应该抛出异常...