php教程

mysql_select_db

(PHP 4, PHP 5)

mysql_select_db — 选择 MySQL 数据库

说明

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

成功时返回 TRUE， 或者在失败时返回 FALSE。

mysql_select_db() 设定与指定的连接标识符所关联的服务器上的当前激活数据库。如果没有指定连接标识符，则使用上一个打开的连接。如果没有打开的连接，本函数将无参数调用 mysql_connect() 来尝试打开一个并使用之。

每个其后的 mysql_query() 调用都会作用于活动数据库。

Example #1 mysql_select_db() 例子

<?php

$lnk = mysql_connect('localhost', 'mysql_user', 'mysql_password')
       or die ('Not connected : ' . mysql_error());

// make foo the current db
mysql_select_db('foo', $lnk) or die ('Can\'t use foo : ' . mysql_error());

?>

参见 mysql_connect()，mysql_pconnect() 和 mysql_query()。

为向下兼容仍然可以使用 mysql_selectdb()，但反对这样做。

参数

database_name

The name of the database that is to be selected.

link_identifier

MySQL 连接。如不指定连接标识，则使用由 mysql_connect() 最近打开的连接。如果没有找到该连接，会尝试不带参数调用 mysql_connect() 来创建。如没有找到连接或无法建立连接，则会生成 E_WARNING 级别的错误。

返回值

成功时返回 TRUE， 或者在失败时返回 FALSE。

范例

Example #2 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());
}
?>

注释

Note:

为了向下兼容，可以使用下列已废弃的别名： mysql_selectdb()

参见

User Contributed Notes

Anonymous 03-Aug-2013 08:22
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();
    }
}
miloshio at gmail dot com 05-May-2012 05:28
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 20-Apr-2012 02:11
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.
riad93 at mail dot ru 12-Sep-2009 04:44
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 04:57
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 08-Sep-2007 11:03
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 05:09
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.
james at gogo dot co dot nz 16-Jan-2004 04:45
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.

CopyRight © 2008-2022 verySource.Com All Rights reserved.　京ICP备17048824号-1　京公网安备：11010502034788