致命错误:在非对象上调用成员函数 fetch_assoc()
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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