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mysql_set_charset> <mysql_result
Last updated: Fri, 10 Jul 2009

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mysql_select_db

(PHP 4, PHP 5)

mysql_select_dbSelect a MySQL database

Descrierea

bool mysql_select_db ( string $database_name [, resource $link_identifier ] )

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.

Parametri

database_name

The name of the database that is to be selected.

link_identifier

Conexiunea MySQL. Dacă identificatorul legăturii nu este specificat, se presupune că este ultima legătură deschisă cu ajutorul mysql_connect(). Dacă nu este găsită nici o astfel de legătură, se va încerca crearea uneia prin apelul mysql_connect() fără argumente. În caz că nici o conexiune nu este găsită sau stabilită, se va genera o eroare de nivelul E_WARNING.

Valorile întroarse

Întoarce valoarea TRUE în cazul succesului sau FALSE în cazul eşecului.

Exemple

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());
}
?>

Note

Notă: Pentru compatibilitatea cu versiunile anterioare, următorul pseudonim dezaprobat poate fi utilizat: mysql_selectdb()

Vedeţi de asemenea



mysql_set_charset> <mysql_result
Last updated: Fri, 10 Jul 2009
 
add a note add a note User Contributed Notes
mysql_select_db
anotheruser at example dot com
12-Aug-2008 11:57
Cross-database join queries, expanding on Dan Ross's post...

Really, this is a mysql specific feature, but worth noting here.  So long as the mysql user has been given the right permissions to all databases and tables where data is pulled from or pushed to, this will work.  Though the mysql_select_db function selects one database, the mysql statement may reference another (the syntax for referencing a field in another db table being 'database.table.field').

<?php

$sql_statement
= "SELECT
    PostID,
    AuthorID,
    Users.tblUsers.Username
    FROM tblPosts
    LEFT JOIN Users.tblUsers ON AuthorID = Users.tblUsers.UserID
    GROUP BY PostID,AuthorID,Username
    "
;

$dblink = mysql_connect("somehost", "someuser", "password");
mysql_select_db("BlogPosts",$dblink);
$qry = mysql_query($sql_statement,$dblink);

?>
me at khurshid dot com
09-Sep-2007 06:03
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.
Maarten
19-Aug-2005 12:09
About opening connections if the same parameters to mysql_connect() are used: this 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.
buzz at oska dot com
06-May-2005 12:39
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...
james at gogo dot co dot nz
17-Jan-2004 12:45
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.
doug at xamo dot com
17-Dec-2003 08:39
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".

<?php
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'");
?>

mysql_set_charset> <mysql_result
Last updated: Fri, 10 Jul 2009
 
 
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