Home Backend Development PHP Tutorial Warning: mysql_close(): 7 is not a valid MySQL-Link resource in,该如何处理

Warning: mysql_close(): 7 is not a valid MySQL-Link resource in,该如何处理

Jun 13, 2016 pm 01:34 PM
gt mysql nbsp quot this

Warning: mysql_close(): 7 is not a valid MySQL-Link resource in
我在一个界面里面进行了两次mysql数据库操作.
1.界面布局分成左右两块,进入界面的时候显示左侧内容,进行一次数据库操作。
2.点击左侧的submit,右侧内容显示,有进行一次数据库操作。
我有一个数据库类,在左边new一次,在右侧new一次。
进入页面就会出现Warning: mysql_close(): 7 is not a valid MySQL-Link resource in E:\www\bk_hmsystem\backstage\db.php on line 22 
如果我删除左侧的数据库操作就不会出现这个问题。
mysql_close放在数据库操作类的析构函数中,如下:
class db
{
  private $host;
  private $user;
  private $pw;
  private $con;
  private $dbname;
   
  function __construct($host,$user,$pw,$dbname)
  {
  $this->host=$host;
  $this->user=$user;
  $this->pw=$pw;
  $this->dbname=$dbname;
  $this->connect();
  }
  function connect()
  {
  $this->con=mysql_connect($this->host,$this->user,$this->pw); 
  if(!$this->con) die(mysql_error()); 
  mysql_select_db($this->dbname,$this->con) or die(mysql_error());
  }
  function __destruct()
  {
  mysql_close($this->con);
  }
  function query($name,$table,$cond)
  {
  if(($name=="")&&($cond==""))
  $sql="select * from $table";
  else 
  {
  if($name=="")
  $sql="select * from $table where $cond";
  else
  {
  if($cond=="")
  $sql="select $name from $table";
  else
  $sql="select $name from $table where $cond";
  }
  }
  $result=mysql_query($sql,$this->con);
  if(!$result)
  {
  die (mysql_error());
  }
  else
  return $result;
  }
}
红色的地方有错.不明白为什么?

------解决方案--------------------
显然调用了两次。第一次已经关闭连接。第二次就会出错。你干脆就省去那一步把。脚本结束后自然会断开连接的。
------解决方案--------------------
你的页面是两个框架页构成的吗?

加些代码分析一下
function __destruct()
{
if(is_resource($this->con))
mysql_close($this->con);
else {
echo $_SERVER['PHP_SELF'];
var_dump($this->con);
}
}

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