php 致命错误：在非对象上调用成员函数 fetch_assoc()
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Fatal error: Call to a member function fetch_assoc() on a non-object
提问by nickles
I'm trying to execute a few queries to get a page of information about some images. I've written a function
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
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;
}
that selects a page of these images based on their popularity. I've written a Databaseclass that handles interactions with the database and an Imageclass that holds information about an image. When I attempt to run this I get an error.
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Database处理与数据库交互的Image类和一个保存图像信息的类。当我尝试运行它时，出现错误。
Fatal error: Call to a member function fetch_assoc() on a non-object
$resultis a mysqli resultset, so I'm baffled as to why this isn't working.
$result是一个 mysqli 结果集，所以我很困惑为什么这不起作用。
回答by ircmaxell
That's because there was an error in your query. MySQli->query()will return false on error. Change it to something like::
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：：
$result = $this->database->query($query);
if (!$result) {
throw new Exception("Database Error [{$this->database->errno}] {$this->database->error}");
}
That should throw an exception if there's an error...
如果有错误，那应该抛出异常......
回答by Marc B
Most likely your query failed, and the query call returned a boolean FALSE (or an error object of some sort), which you then try to use as if was a resultset object, causing the error. Try something like var_dump($result)to see what you really got.
很可能您的查询失败，并且查询调用返回一个布尔值 FALSE（或某种错误对象），然后您尝试将其用作结果集对象，从而导致错误。尝试类似的事情var_dump($result)，看看你真正得到了什么。
Check for errors after EVERY database query call. Even if the query itself is syntactically valid, there's far too many reasons for it to fail anyways - checking for errors every time will save you a lot of grief at some point.
在每次数据库查询调用后检查错误。即使查询本身在语法上是有效的，它失败的原因也太多了 - 每次检查错误都会在某些时候为您节省很多麻烦。
回答by Upendra Bittu
I happen to miss spaces in my query and this error comes.
我碰巧在查询中遗漏了空格，并且出现了此错误。
Ex: $sql= "SELECT * FROM";
$sql .= "table1";
Though the example might look simple, when coding complex queries, the probability for this error is high. I was missing space before word "table1".
尽管该示例可能看起来很简单，但在编码复杂查询时，出现此错误的可能性很高。我在单词“table1”之前缺少空格。
回答by Mustafizur Rahman
Please check if you have already close the database connection or not. In my case i was getting the error because the connection was close in upper line.
请检查您是否已经关闭了数据库连接。在我的情况下，我收到错误，因为连接在上线关闭。
