Connect-phpmysql connection problem. Can't two connections be generated at the same time.

A MySql class {code...} is encapsulated, And the MySql connection is instantiated in the two classes and the data is obtained by defining methods. {Code...} and then create a new test. php page. If you reference any UserDao or NoteDao separately, there is no problem. However, if a MySql class is encapsulated at the same time...

class MySql{    private $dbCon;    public function __construct(){        $this->dbCon=mysql_connect(...);        mysql_select_db("test",$this->dbCon);    }    public function query($query){        return mysql_query($query,$this->dbCon);    }    public function closeDb(){        mysql_close($this->dbCon);    }}

In addition, the MySql connection is instantiated in two classes and the method is defined to obtain data.

class UserDao{    private $mysql;    public function __construct(){       $this->mysql=new MySql();    }    public function getUsers(){      $query="SELECT * FROM USERS";      $rs=$this->mysql->query($query);      #format data      return $result;    }    public function __destruct(){        $this->mysql->closeDb();    }}

class NoteDao{    private $mysql;    public function __construct(){       $this->mysql=new MySql();    }    public function getNotes(){      $query="SELECT * FROM Notes";      $rs=$this->mysql->query($query);      #format data      return $result;    }    public function __destruct(){        $this->mysql->closeDb();    }}

Create a test. php page
If you reference any UserDao or NoteDao separately, there is no problem. However, if both UserDao and NoteDao are referenced to obtain data
PHP Warning: mysql_close (): 9 is not a valid MySQL-Link resource in...
This error occurs.
Why?

Reply content:

A MySql class is encapsulated.

class MySql{    private $dbCon;    public function __construct(){        $this->dbCon=mysql_connect(...);        mysql_select_db("test",$this->dbCon);    }    public function query($query){        return mysql_query($query,$this->dbCon);    }    public function closeDb(){        mysql_close($this->dbCon);    }}

In addition, the MySql connection is instantiated in two classes and the method is defined to obtain data.

class UserDao{    private $mysql;    public function __construct(){       $this->mysql=new MySql();    }    public function getUsers(){      $query="SELECT * FROM USERS";      $rs=$this->mysql->query($query);      #format data      return $result;    }    public function __destruct(){        $this->mysql->closeDb();    }}

class NoteDao{    private $mysql;    public function __construct(){       $this->mysql=new MySql();    }    public function getNotes(){      $query="SELECT * FROM Notes";      $rs=$this->mysql->query($query);      #format data      return $result;    }    public function __destruct(){        $this->mysql->closeDb();    }}

Create a test. php page
If you reference any UserDao or NoteDao separately, there is no problem. However, if both UserDao and NoteDao are referenced to obtain data
PHP Warning: mysql_close (): 9 is not a valid MySQL-Link resource in...
This error occurs.
Why?

Mysql_connect will reuse the connection. Therefore, the two new MySql instances use the same connection (you can see var_dump that the resource id is the same ).
The fourth parameter of mysql_connect is whether to create a new connection. The default value is FALSE. Set this parameter to TRUE.

$this->dbCon = mysql_connect($host, $user, $passwd, TRUE);

However, I do not recommend that you do this. connections can be reused ~~~ No need to create. The connection to php created with mysql_connect is automatically released without calling mysql_colse.

Code written incorrectly

$dbCon=mysql_connect(...);

It should be

$this->dbCon=mysql_connect(...);