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.
mysql_select_db
(PHP 4, PHP 5)
mysql_select_db — Sélectionne une base de données MySQL
Avertissement
Cette extension était obsolète en PHP 5.5.0, et a été supprimée en PHP 7.0.0. À la place, vous pouvez utiliser l'extension MySQLi ou l'extension PDO_MySQL. Voir aussi MySQL : choisir une API du guide. Alternatives à cette fonction :
- mysqli_select_db()
- PDO::__construct() (fait parti du dsn)
Description
mysql_select_db(string
$database_name, resource $link_identifier = NULL): bool
Sélectionne une base de données MySQL sur le serveur associé avec le paramètre
link_identifier. Chaque appel à la fonction
mysql_query() sera exécutée sur la base de données active.
Liste de paramètres
-
database_name -
Le nom de la base de données à sélectionner.
-
link_identifier -
La connexion MySQL. S'il n'est pas spécifié, la dernière connexion ouverte avec la fonction mysql_connect() sera utilisée. Si une telle connexion n'est pas trouvée, la fonction tentera d'ouvrir une connexion, comme si la fonction mysql_connect() avait été appelée sans argument. Si aucune connexion n'est trouvée ou établie, une alerte
E_WARNINGest générée.
Valeurs de retour
Cette fonction retourne true en cas de succès ou false si une erreur survient.
Exemples
Exemple #1 Exemple avec mysql_select_db()
<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Impossible de se connecter : ' . mysql_error());
}
// Rendre la base de données foo, la base courante
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
die ('Impossible de sélectionner la base de données : ' . mysql_error());
}
?>
Notes
Note:
Pour des raisons de compatibilité ascendante, l'alias obsolète suivant peut être utilisé : mysql_selectdb()
Voir aussi
- mysql_connect() - Ouvre une connexion à un serveur MySQL
- mysql_pconnect() - Ouvre une connexion persistante à un serveur MySQL
- mysql_query() - Envoie une requête à un serveur MySQL
add a note
User Contributed Notes 8 notes
james at gogo dot co dot nz ¶
20 years ago
Maarten ¶
18 years ago
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.
Anonymous ¶
10 years ago
function go_mysql($query)
{
global $mysql_link;
if (!$mysql_link)
{
$mysql_link = mysql_connect("localhost","root","my_pass") or die(mysql_error());
mysql_select_db("my_db") or die(mysql_error());
mysql_query("SET NAMES 'utf8'");
mysql_query("set character_set_client='utf8'");
mysql_query("set character_set_results='utf8'");
mysql_query("set collation_connection='utf8'");
global $mysql_link;
}
$result=mysql_query($query);
if ($result)
{
return $result;
}
else
{
echo "Database Error: " . mysql_error()."<br><b>$query</b>";
die();
}
}
anotheruser at example dot com ¶
15 years ago
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);
?>
miloshio at gmail dot com ¶
11 years ago
You can select MySQL database without using this function.
Simply right after connecting to MySQL
<?php $connection = mysql_connect("dabatbasehost", "username", "password"); ?>
perform this query:
<?php mysql_query("USE somedatabase", $connection); ?>
duncan at berrimans dot co dot uk ¶
12 years ago
Note that the manual is slightly misleading it states :-
"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."
The 2nd statement is not true or at best unclear.
mysql_query() manual entry actually correctly states it will use the last link opened by mysql_connect() by default.
Thus if you have 2 connections you will need to specify the connection when calling mysql_query or issue the connect again to ensure the 1st database becomes the default, simply using mysql_select_db will not make the 1st database the default for subsequent calls to mysql_query.
Its probably only apparent when the two databases are on different servers.
me at khurshid dot com ¶
16 years ago
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.
riad93 at mail dot ru ¶
14 years ago
You can use DataBases without <?php mysql_select_db() ?>
And you will havenot james at gogo dot co dot nz's problems :)
<?php
mysql_connect('localhost','db_user','pssword');
mysql_query('SELECT * FROM database_name.table_name');
?>
