我在运行的网站上有以下错误。我不明白为什么它在我的localhost上工作正常。这与主持人有关吗?我在Unix服务器上。
Warning: mysqli::mysqli() [mysqli.mysqli]: (42000/1203): User dbo343879423 already has more than 'max_user_connections' active connections in /homepages/9/d322397966/htdocs/dump/models/class_database.php on line 11
Connect failed: User dbo343879423 already has more than 'max_user_connections' active connections
Warning: mysqli::close() [mysqli.close]: Couldn't fetch mysqli in /homepages/9/d322397966/htdocs/dump/models/class_database.php on line 160
错误显示'用户dbo343879423已经在第11行的/homepages/9/d322397966/htdocs/dump/models/class_database.php中有超过'max_user_connections'活动连接,所以这是脚本中的第11行 - 我看不出任何错误!
$this -> connection = new mysqli($hostname,$username,$password,$database);
下面是class_database.php中的整个类,在脚本的其他部分是否有错,我应该更改?
<?php
#connects the database and handling the result
class __database {
protected $connection = null;
protected $error = null;
#make a connection
public function __construct($hostname,$username,$password,$database)
{
$this -> connection = new mysqli($hostname,$username,$password,$database);
if (mysqli_connect_errno())
{
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
}
#fetches all result rows as an associative array, a numeric array, or both
public function fetch_all($query)
{
$result = $this -> connection -> query($query);
if($result)
{
return $result -> fetch_all(MYSQLI_ASSOC);
}
else
{
$this -> error = $this -> connection -> error;
return false;
}
}
#fetches a result row as an associative array, a numeric array, or both
public function fetch_assoc_while($query)
{
$result = $this -> connection -> query($query);
if($result)
{
while($row = $result -> fetch_assoc())
{
$return_this[] = $row;
}
if (isset($return_this))
{
return $return_this;
}
else
{
return false;
}
}
else
{
$this -> error = $this -> connection -> error;
return false;
}
}
#fetch a result row as an associative array
public function fetch_assoc($query)
{
$result = $this -> connection -> query($query);
if($result)
{
return $result -> fetch_assoc();
}
else
{
$this -> error = $this -> connection -> error;
return false;
}
}
#get a result row as an enumerated array
public function fetch_row($query)
{
$result = $this -> connection -> query($query);
if($result)
{
return $result -> fetch_row();
}
else
{
$this -> error = $this -> connection -> error;
return false;
}
}
#get the number of rows in a result
public function num_rows($query)
{
$result = $this -> connection -> query($query);
if($result)
{
return $result -> num_rows;
}
else
{
$this -> error = $this -> connection -> error;
return false;
}
}
#performs a query on the database
public function query($query)
{
$result = $this -> connection -> query($query);
if($result)
{
return $result;
}
else
{
$this -> error = $this -> connection -> error;
return false;
}
}
#escapes special characters in a string for use in a SQL statement, taking into account the current charset of the connection
public function real_escape_string($string)
{
$result = $this -> connection -> real_escape_string($string);
if($result)
{
return $result;
}
else
{
$this -> error = $this -> connection -> error;
return false;
}
}
#display error
public function get_error()
{
return $this -> error;
}
#closes the database connection when object is destroyed.
public function __destruct()
{
$this -> connection -> close();
}
}
?>
或者我应该只是改变主机!??
下面是数据库连接类的实现。如果我把这部分拿出来,错误将不再出现,但我也会在网站的其他部分也这样做,它们不会造成任何问题!
<!-- side-video-library -->
<div id="side-video-library" class="round-corner">
<h4><a href="<?php echo HTTP_ROOT;?>videos"><span>ENER VIDEO LIBRARY</span></a></h4>
<?php
$sql = "
SELECT *
FROM root_pages
WHERE root_pages.parent_id = '8'
AND root_pages.pg_highlight = '1'
AND root_pages.pg_hide != '1'
ORDER BY rand() DESC
LIMIT 1
";
#instantiate the object of __database class
$object_item = new __database(DB_HOST,DB_USER,DB_PASS,DB_NAME);
$item = $object_item -> fetch_assoc($sql);
#instantiate the object of __database class
$object_item_num = new __database(DB_HOST,DB_USER,DB_PASS,DB_NAME);
$total_item = $object_item_num -> num_rows($sql);
//echo $total_item;
?>
<?php
if ($total_item > 0)
{
$sql = "
SELECT *
FROM root_tagged
LEFT JOIN root_tags ON ( root_tags.tag_id = root_tagged.tag_id )
WHERE root_tagged.pg_id = '".$item['pg_id']."'
";
#instantiate the object of __database class
$object_tagname = new __database(DB_HOST,DB_USER,DB_PASS,DB_NAME);
$item_tagname = $object_tagname -> fetch_assoc($sql);
#instantiate the object of __database class
$object_tagname_num = new __database(DB_HOST,DB_USER,DB_PASS,DB_NAME);
$total_tagname = $object_tagname_num -> num_rows($sql);
?>
<p class="item-video">
<object style="width: 183px; height: 151px;" width="183" height="151" data="http://www.youtube.com/v/<?php echo get_video_id($item['pg_content_1']) ;?>" type="application/x-shockwave-flash">
<param name="wmode" value="transparent" />
<param name="src" value="http://www.youtube.com/v/<?php echo get_video_id($item['pg_content_1']) ;?>" />
</object>
</p>
<h3><a href="<?php echo HTTP_ROOT.str_replace(' ', '-', 'videos').'/'.$item_tagname['tag_name'].'/'.str_replace(' ', '-', strtolower($item['pg_url']));?>"><?php if(strlen($item['pg_title']) > 20) echo substr($item['pg_title'], 0,20).'...'; else echo $item['pg_title'];?></a></h3>
<p class="item-excerpt-video"><?php if(strlen($item['pg_content_2']) > 100) echo substr($item['pg_content_2'], 0,100).'...'; else echo $item['pg_content_2'];?></p>
<a href="<?php echo HTTP_ROOT;?>videos" class="button-arrow"><span>More</span></a>
<?php
}
?>
</div>
<!-- side-video-library -->
我是否一直在错误地实施课程?
感谢。
答案 0 :(得分:10)
问题可能是您只允许少量连接,当您的类尝试获取新连接时,您会遇到此错误。
这不是编程问题,只是可用资源的数量。使用此类的任何其他脚本都有错误。
您必须在服务器上的mysql配置文件上配置更多连接。如果您没有此访问权限,请要求支持人员进行此操作,或者更改为允许更多连接的托管公司!
其他选项是在此类上实现Singleton模式,因此它重用相同的连接池,并且不会爆炸限制。
答案 1 :(得分:5)
如果您收到此max_user_connections消息,请先优化数据库表。
如何优化数据库表和查询:
答案 2 :(得分:3)
检查MySQL服务器上的MAX USER_CONNECTIONS设置以供用户使用。在PHPMyAdmin中,转到服务器页面(单击Server:&lt;&gt;),然后在子菜单中单击priviledges。编辑用户dbo343879423,MAX USER_CONNECTIONS将在右侧。默认情况下,我认为它设置为0(无限制),您的可能会受到限制,具体取决于谁设置服务器。
我不确定您的Database类是如何使用的,但如果您多次实例化该类,请考虑在数据库类中创建一个私有静态变量Database并创建一个公共静态方法getDatabase(),它实例化数据库连接,如果它为null并返回实例。
答案 3 :(得分:1)
为了它的价值,我想要包含一个情况,我遇到了由于占位符不正确而收到此消息的情况:
$sql_str = 'SELECT prod_id FROM ' . $this->table_name["product"] . ' WHERE prod_sku =:p_sku';
$arr[':prod_sku'] = $s_sku;
另外我做了大量的查询。我怀疑错误与大量查询相结合导致了这个问题。当我修复查询时,连接会在离开时发出。
答案 4 :(得分:1)
对于Godaddy共享主机,您无法更改MAX_USER_CONNECTION值。要找到它,请单击服务器&lt;&gt;,然后单击菜单栏中的变量。我的设定为200。
答案 5 :(得分:0)
使用命令行检查登录到mysql的连接数并运行此命令
SHOW VARIABLES LIKE 'max_user_connections
步骤1:在/etc/my.cnf中查看设置
max_user_connections = <set number that you want>
max_user_connections =
您也可以从mysql命令行设置
SET GLOBAL max_user_connections = 0;
重新启动apache服务器