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Home/ Questions/Q 687191
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T02:06:06+00:00 2026-05-14T02:06:06+00:00

I have an application in which I want to authenticate a user from a

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I have an application in which I want to authenticate a user from a first database & manage other activities from another database.
I have created two classes. An object of the classes is defined in a file:

$objdb1=new db1(),$objdb2=new db2();

But when I try to call $objdb1->fn(). It searches from the $objdb2 & is showing table1 doesnot exists?

My first file database.php

class database
{
private $hostname;
private $database;
private $username;
private $password;
private $dblinkid;

function __construct()
{
    if($_SERVER['SERVER_NAME'] == 'localhost')
    {
        $this->hostname = "localhost";
        $this->database = "aaaa";
        $this->username = "xxx";
        $this->password = "";
    }
    else
    {
        $this->hostname = "localhost";
        $this->database = "xxx";
        $this->username = "xxx";
        $this->password = "xxx";
    }
    $this->dblinkid = $this->connect();
}

       protected function connect()
{
    $linkid = mysql_connect($this->hostname, $this->username, $this->password) or die("Could not Connect ".mysql_errno($linkid));
    mysql_select_db($this->database, $linkid) or die("Could not select database ".mysql_errno($linkid)) ;
    return $linkid;
}

Similarly second file

class database2
{
private $vhostname;
private $vdatabase;
private $vusername;
private $vpassword;
private $vdblinkid;

function __construct()
{
    if($_SERVER['SERVER_NAME'] == 'localhost')
    {
        $this->vhostname = "xxx";
        $this->vdatabase = "bbbb";
        $this->vusername = "xxx";
        $this->vpassword = "";
    }
    else
    {
        $this->vhostname = "localhost";
        $this->vdatabase = "xxxx";
        $this->vusername = "xxxx";
        $this->vpassword = "xxxx";
    }
    $this->vdblinkid = $this->vconnect();
}

        protected function vconnect()
{
    $vlinkid = mysql_connect($this->vhostname, $this->vusername, $this->vpassword) or die("Could not Connect ".mysql_errno($vlinkid));
    mysql_select_db($this->vdatabase, $vlinkid) or die("Could not select database ".mysql_errno($vlinkid)) ;
    return $vlinkid;
}

Third file

$objdb1 = new database();
$objdb2 = new database2();

Can you help me on this?

Regards,

Pankaj

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T02:06:06+00:00Added an answer on May 14, 2026 at 2:06 am

    Without knowing your classes, it is difficult to help. If you are using PDO, I can guarantee you that you can create multiple instances connected to different databases without any problem. If you are using the mysql_ family of functions you probably just forgot to set the link_identifier parameter (see here).

    However, having a class db1 and a class db2 sounds like a code smell to me. You probably want to have two instances of the same class with different attributes.

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