Calling two databases in one login form

[Craziest]

i need to have two databases while am login. one fake and one original . if
the fake login and the password is true then it will redirect him to another
page. so i need to have two require statements so it can check both databas
es,,
if its possible tht would be great...
i will be greatful for any comments....

[ronverdonk]

Following is a quote from an entry in the php.net documentation for mysql_select_db:

[quote]Originally Posted by
matsko at rogers dot com 10-May-2006 09:19
Just incase the mysql_select_db() function still won't work with multiple database connections (as has happened to me before).

$dbh1 = mysql_pconnect($host,$user,$pass);
$dbh2 = mysql_pconnect($host,$user,$pass);

You could do this...

mysql_query("USE database1",$dbh1);
mysql_query("Use database2",$dbh2);

This does the same thing as the mysql_select_db() function...

or this...

You don't even have to select the database for each connection.

mysql_query("SELECT * FROM database1.table",$dbh1);
mysql_query("SELECT * FROM database2.table",$dbh2);
[/quote]

[redarrow]

[code]
<?php

if($name && $password){

if($_POST['submit']) {

$db1=mysql_connect("xxxx","xxxxx","xxxxx");

$result1=mysql_select_db("xxxxx",$db1);

$query1="select * from  xxxxx where id='$id'":

$resulted1=mysql_query($query1);

if(mysql_num_rows($resulted1)){

echo " your logged into database 1";

}else{

$db2=mysql_connect("xxxx","xxxxx","xxxxx");

$result2=mysql_select_db("xxxxx",$db2);

$query2="select * from xxxxxxx where id='$id'":

$resulted2=mysql_query($query2);

if(mysql_num_rows($resulted2)){

echo " your logged into database 2";
}
}
  }else{

echo " Sorry but we have no deatail of your account please register thank you.";

include("register.php");

}
?>
[/code]

[Craziest]

Thanks guys for the quick reply,im still getting afew problems since im using wordpress and adding this feature to the login form.
this is what i added
wp-login.php
[code]<?php

require('./wp-config.php');

function login($username, $password, $already_md5 = false) {

global $wpdb, $wpdb1, $error, $tableusers;
if($username && $password){

if($_POST['submit']) {

$wpdb=mysql_connect($server,$loginsql,$passsql);

$query = "SELECT ID, user_login, user_pass FROM $tableusers, WHERE user_login = '$username'";

$login =mysql_query($query);

}else{

$wpdb2=mysql_connect($server1,$loginsql1,$passsql1);

$query2 = "SELECT ID, user_login, user_pass FROM $tableusers, WHERE user_login = '$username'";

$login =mysql_query($query2);[/code]

and in the wp-config.php
[code]// ** MySQL settings ** //
define('DB_NAME', 'wp1');    // The name of the database
define('DB_USER', 'root');    // Your MySQL username
define('DB_PASSWORD', ''); // ...and password
define('DB_HOST', 'localhost');    // 99% chance you won't need to change this value

define('DB_NAME1', 'wp1fake');    // The name of the database
define('DB_USER1', 'root');    // Your MySQL username
define('DB_PASSWORD1', ''); // ...and password
define('DB_HOST1', 'localhost');    // 99% chance you won't need to change this value

$table_prefix  = 'wp_';  // example: 'wp_' or 'b2' or 'mylogin_'

define ('WPLANG', '');

/* Stop editing */

$server = DB_HOST;
$loginsql = DB_USER;
$passsql = DB_PASSWORD;
$base = DB_NAME;

/* Stop editing */

$server1 = DB_HOST1;
$loginsql1 = DB_USER1;
$passsql1 = DB_PASSWORD1;
$base1 = DB_NAME1;
[/code]
When i enter the username and pass in wp-login.php it says
Warning: mysql_connect(): Access denied for user 'ODBC'@'localhost' (using password: NO) in c:\appserv\www\wp\wp-login.php on line 14
which is [code]$query = "SELECT ID, user_login, user_pass FROM $tableusers, WHERE user_login = '$username'";[/code]
Sorry im brand new in php :(  please excuse my silly mistakes.

[bubazoo]

huh, .....and I always thought u had to do a join  (shrugs)  ...learn something new everyday I suppose