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POConnecting to two databases
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copied!<p>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:</p> <pre><code>$objdb1=new db1(),$objdb2=new db2(); </code></pre> <p>But when I try to call <code>$objdb1->fn()</code>. It searches from the <code>$objdb2</code> & is showing table1 doesnot exists?</p> <p>My first file database.php</p> <pre><code>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; } </code></pre> <p>Similarly second file</p> <pre><code>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 $vlinkid; } </code></pre> <p>Third file</p> <pre><code>$objdb1 = new database(); $objdb2 = new database2(); </code></pre> <p>Can you help me on this?</p> <p>Regards,</p> <p>Pankaj</p>

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