我正在运行的网站上出现以下错误。我不明白为什么会这样,因为它在我的本地主机上可以正常工作。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在第11行的/homepages/9/d322397966/htdocs/dump/models/class_database.php中已经具有超过’max_user_connections’个活动连接”,所以这是脚本中的第11行-我可以。看不到任何错误!
$this -> connection = new mysqli($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出错,我应该更改吗?
<?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(); } } ?>
还是我应该永久更改主机!?
下面是数据库连接类的实现。如果我删除此部分,该错误将不再出现,但是我在网站的其他部分也进行了同样的操作,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 -->
我上课有误吗?
谢谢。
可能的问题是您只允许少数几个连接,并且当您的班级尝试获取新的连接时,会出现此错误。
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均会发生错误。
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更改为允许更多连接的托管公司!
另一种选择是在此类上实现单例模式,因此它可以重用相同的连接池,并且不会超出限制。