致命错误:在非对象上调用成员函数 fetch_assoc() [英] Fatal error: Call to a member function fetch_assoc() on a non-object

问题描述

我正在尝试执行一些查询以获取有关某些图像的信息页面.我写了一个函数

I'm trying to execute a few queries to get a page of information about some images. I've written a function

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;
}

根据它们的受欢迎程度选择这些图像的页面.我编写了一个处理与数据库交互的 Database 类和一个保存图像信息的 Image 类.当我尝试运行此程序时出现错误.

that selects a page of these images based on their popularity. I've written a Database class that handles interactions with the database and an Image class that holds information about an image. When I attempt to run this I get an error.

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.

$result is a mysqli resultset, so I'm baffled as to why this isn't working.

推荐答案

那是因为您的查询中有错误.MySQli->query()出错时返回false.将其更改为类似::

That's because there was an error in your query. MySQli->query() will return false on error. Change it to something like::

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

如果有错误应该抛出异常...

That should throw an exception if there's an error...

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