mysql_select_db

(PHP 4, PHP 5)

mysql_select_db — MySQL データベースを選択する

説明

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

指定したリンク ID が指すサーバ上のデータベースを、アクティブなデータベースに設定します。それ以降にコールされる mysql_query() は、すべてアクティブなデータベース上で実行されます。

パラメータ

database_name: 選択するデータベース名。
link_identifier: MySQL 接続。指定されない場合、mysql_connect() により直近にオープンされたリンクが指定されたと仮定されます。そのようなリンクがない場合、引数を指定せずに mysql_connect() がコールした時と同様にリンクを確立します。リンクが見付からない、または、確立できない場合、 E_WARNING レベルのエラーが生成されます。

返り値

成功した場合に TRUE を、失敗した場合に FALSE を返します。

例

例1 mysql_select_db() の例


<?php

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Not connected : ' . mysql_error());
}

// foo をカレントの db に指定する
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
    die ('Can\'t use foo : ' . mysql_error());
}
?>

注意

注意: 下位互換のために、次の非推奨別名を使用してもいいでしょう。 mysql_selectdb()

参考

mysql_connect() - MySQL サーバへの接続をオープンする
mysql_pconnect() - MySQL サーバへの持続的な接続をオープンする
mysql_query() - MySQL クエリを送信する

mysql_set_charset

mysql_result

Last updated: Fri, 13 Nov 2009

add a note User Contributed Notes
mysql_select_db

riad93 at mail dot ru
12-Sep-2009 11:44


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');



?>

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'");


?>

add a note

mysql_set_charset

mysql_result

Last updated: Fri, 13 Nov 2009