just to improve buzz at oska dot com solution, it is definitely the best solution so far
but it works only if you use the new_link parameter in the mysql_connect, setting it to true
be aware that the new_link parameter was introduced starting from PHP 4.2
if you don't set the new_link to true, you will need to specify the database name followed by table name in the FROM query, ie. FROM databaseName.tableName, and probably in the SELECT
more information here
http://uk.php.net/function.mysql-connect
mysql_select_db
(PHP 4, PHP 5, PECL mysql:1.0)
mysql_select_db — Select a MySQL database
Description
Sets the current active database on the server that's associated with the specified link identifier. Every subsequent call to mysql_query() will be made on the active database.
Parameters
- database_name
-
The name of the database that is to be selected.
- link_identifier
-
The MySQL connection. If the link identifier is not specified, the last link opened by mysql_connect() is assumed. If no such link is found, it will try to create one as if mysql_connect() was called with no arguments. If by chance no connection is found or established, an E_WARNING level error is generated.
Return Values
Returns TRUE on success or FALSE on failure.
Examples
Example #1 mysql_select_db() example
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Not connected : ' . mysql_error());
}
// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
?>
Notes
Note: For backward compatibility, the following deprecated alias may be used: mysql_selectdb()
mysql_select_db
21-May-2008 06:39
09-Sep-2007 03:03
Problem with connecting to multiple databases within the same server is that every time you do:
mysql_connect(host, username, passwd);
it will reuse 'Resource id' for every connection, which means you will end with only one connection reference to avoid that do:
mysql_connect(host, username, passwd, true);
keeps all connections separate.
11-May-2006 04:19
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);
19-Aug-2005 09:09
Previously posted comments about opening connections if the same parameters to mysql_connect() are used can be avoided by using the 'new_link' parameter to that function.
This parameter has been available since PHP 4.2.0 and allows you to open a new link even if the call uses the same parameters.
06-May-2005 09:39
As has been already commented, opening multiple connection handles with:
<?php
$connection_handle = mysql_connect($hostname_and_port,$user,$password);
?>
causes the connection ID/handle to be REUSED if the exact same parameters are passed in to it. This can be annoying if you want to work with multiple databases on the same server, but don't want to (a) use the database.table syntax in all your queries or (b) call the mysql_select_db($database) before every SQL query just to be sure which database you are working with.
My solution is to create a handle for each database with mysql_connect (using ever so slightly different connection properties), and assign each of them to their own database permanently. each time I do a mysql_query(...) call, I just include the connection handle that I want to do this call on eg (ive left out all error checking for simplicity sake):
<?php
// none of thesehandles are re-used as the connection parameters are different on them all, despite connecting to the same server (assuming 'myuser' and 'otheruser' have the same privileges/accesses in mysql)
$handle_db1 = mysql_connect("localhost","myuser","apasswd");
$handle_db2 = mysql_connect("127.0.0.1","myuser","apasswd");
$handle_db3 = mysql_connect("localhost:3306","myuser","apasswd");
$handle_db4 = mysql_connect("localhost","otheruser","apasswd");
// give each handle it's own database to work with, permanently.
mysql_select_db("db1",$handle_db1);
mysql_select_db("db2",$handle_db2);
mysql_select_db("db3",$handle_db3);
mysql_select_db("db4",$handle_db4);
//do a query from db1:
$query = "select * from test"; $which = $handle_db1;
mysql_query($query,$which);
//do a query from db2 :
$query = "select * from test"; $which = $handle_db2;
mysql_query($query,$which);
//etc
?>
Note that we didn't do a mysql_select_db between queries , and we didn't use the database name in the query either.
Of course, it has the overhead of setting up an extra connection.... but you may find this is preferable in some cases...
13-Feb-2004 06:43
Another way to select from 2 different databases on the same server:
mysql_select_db("db1");
$res_db1 = mysql_query("select * from db1.foobar");
$res_db2 = mysql_query("select * from db2.foobar");
I.e. just prepend database name.
17-Jan-2004 09:45
Be carefull if you are using two databases on the same server at the same time. By default mysql_connect returns the same connection ID for multiple calls with the same server parameters, which means if you do
<?php
$db1 = mysql_connect(...stuff...);
$db2 = mysql_connect(...stuff...);
mysql_select_db('db1', $db1);
mysql_select_db('db2', $db2);
?>
then $db1 will actually have selected the database 'db2', because the second call to mysql_connect just returned the already opened connection ID !
You have two options here, eiher you have to call mysql_select_db before each query you do, or if you're using php4.2+ there is a parameter to mysql_connect to force the creation of a new link.
18-Dec-2003 05:39
When you need to query data from multiple databases, note that mysql_select_db("db2") doesn't prevent you from fetching more rows with result sets returned from "db1".
mysql_select_db("db1");
$res_db1=mysql_query("select * from foobar");
myqsl_select_db("db2);
$row_db1=mysql_fetch_object($res_db1);
$res_db2=mysql_query("select * from test where id='$row_db1->id'");
