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mysql_select_db (PHP 3, PHP 4 ) mysql_select_db -- Sélectionne une base de données MySQL Descriptionbool mysql_select_db ( string database_name [, resource link_identifier])
mysql_select_db() change la base de données
active sur la connexion représentée par
link_identifier. Si aucun identifiant
n'est spécifié, la
dernière connexion est utilisée. S'il n'y a pas de
dernière connexion, la fonction tentera de se connecter seule, avec
mysql_connect() et les paramètres par
défaut.
Cette fonction retourne TRUE en cas de succès, FALSE en cas d'échec.
Toutes les requêtes suivantes avec mysql_query()
seront faites avec la base de données active.
Voir aussi
mysql_connect(),
mysql_pconnect() et
mysql_query().
Pour des raisons de compatibilité ascendante mysql_selectdb() est
encore disponible. Elle est toutefois obsolète.
User Contributed Notes
mysql_select_db |
 add a note |
DEF2CEo${#]KK?]4@>
04-Jun-2003 04:57 |
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Oromian at Hotmail offers some "favorite" code but it uses the
deprecated mysql_db_query function and offers nothing to those looking for
help with mysql_select_db(). Under mysql_db_query() is the
following:
Note: This function has been deprecated since PHP 4.0.6.
Do not use this function. Use mysql_select_db() and mysql_query()
instead.
(email addr is ROT47 encoded)
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alphalund at infinito dot it
06-May-2003 02:16 |
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It seems that mySQL doesn't like the backtick (`) operator when changing
DB.
So if you call
mysql_select_db('`some-db`',$link) ;
you will
get an error.
However the same request will be successful if you
delete backticks
mysql_select_db('some-db',$link);
Note that
it's a mySql problem. The same problem appears if you try to do the same
thing with a pure SQL statement:
USE DB `some-db`
won't Work! The
bigger problem is that neither
USE DB some-db
will work because of
the '-'
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Oromian at hotmail dot com
30-Mar-2003 06:33 |
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here is my favorite way to do the DB thang. :)
<?
// db
variables
// inserts variables accordingly
$host =
"";
$user = "";
$pass = "";
$db =
"";
$table = "";
// connecting to the
db
$link = mysql_connect ($host, $user, $pass);
// the db
query
$q = "SELECT * from $table";
$results =
mysql_db_query ($db, $q, $link);
// fetching the results
while
($row = mysql_fetch_array($result)){
echo
"$row[whatever]
$row[whatever2]
";
}
/*
this is the area you can grab stuff from the db in the manner i
did it above
whatever = name of the tables colum
*/
// closing the db link
mysql_close
($link);
?>
i'd be happy to answer some questions about this
topic for you newbies our there, just drop me an email.
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isox at chainsawbeer dot com
24-Feb-2003 05:05 |
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Note that if the call to mysql_select_db() fails, mysql_errno() and
mysql_error() are available with the error related information.
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rhcubevn at netdoor dot com
29-Nov-2002 03:45 |
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The previous post was a bad example. Here is a better example, with the
optional resource parm:
<?php
$connection =
mysql_connect($host, $user, $pass);
$db = mysql_select_db($mydatabase,
$connection);
?>
The second paremeter in mysql_select_db is
if you defined more than one mysql_connect(). You should *always* define
it, even if you only have one connection, as it is good practice, and will
prevent confusion later if you want to add another connection.
P.S.
The first part of my e-mail (before @) is rot_13 encoded!
Armond
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jorik at scriptsnet dot nl
14-Sep-2002 05:44 |
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<?PHP
mysql_connect("your host",
"username", "password") or
die(mysql_error());
mysql_select_db("your database") or
die(mysql_error());
?>
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add a note |
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Manuel PHP
MySQL
mysql_affected_rows
mysql_select_db
mysql_result