#43947 [Opn]: mysql_select_db fails to select database using variables
| From: | joel dot a dot villarreal at gmail dot com | Date: | Sun, 27 Jan 2008 22:40:01 +0000 |
| Subject: | #43947 [Opn]: mysql_select_db fails to select database using variables | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-121583@lists.php.net to get a copy of this message | ||
ID: 43947
User updated by: joel dot a dot villarreal at gmail dot com
Reported By: joel dot a dot villarreal at gmail dot com
Status: Open
Bug Type: MySQL related
Operating System: Windows XP Service Pack 2
PHP Version: 5.2.5
New Comment:
Just in case, these are the versions of the server components being
used at the moment:
PHP Version 5.2.5
MySQL 5.0.46
Apaache 2.2.6
Previous Comments:
------------------------------------------------------------------------
[2008-01-27 22:38:14] joel dot a dot villarreal at gmail dot com
Description:
------------
When calling a function to connect to MySQL, while supplying a database
name for mysql_select_db() through a variable defined at an include
file, the database is not selected.
Reproduce code:
---------------
## connection.cfg.php says:
<?
$MySQL_DatabaseName = "foo";
?>
## main.php says:
<?
include("connection.cfg.php");
connect();
function connect() {
global $MySQL_DatabaseName;
mysql_connect("localhost", "root", "########");
mysql_select_db($MySQL_DatabaseName);
$row = mysql_fetch_array(mysql_query("SELECT * FROM footable WHERE ID =
1"), MYSQL_ASSOC) or die(mysql_error());
print_r($row);
}
?>
Expected result:
----------------
Expected result is to obtain a view of the row's data through the
print_r() function.
Actual result:
--------------
mysql_error() prints: "No database selected".
PHP warns mysql_fetch_array() has been given an invalid MySQL result
resource.
------------------------------------------------------------------------
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Edit this bug report at http://bugs.php.net/?id=43947&edit=1