ID: 43947
Comment by: will dot fitch 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:
I can't reproduce this. Here is my code:
connection.cfg.php
<?
$MySQL_DatabaseName = "mysql";
?>
test.php
<?
include("connection.cfg.php");
connect();
function connect() {
global $MySQL_DatabaseName;
mysql_connect("localhost", "root", "########");
mysql_select_db($MySQL_DatabaseName);
$row = mysql_fetch_array(mysql_query("SELECT * FROM user"),
MYSQL_ASSOC) or die(mysql_error());
print_r($row);
}
?>
It prints out exactly as expected. I am using PHP 5.2.5 on RHEL5.
Previous Comments:
------------------------------------------------------------------------
[2008-01-29 17:28:34] joel dot a dot villarreal at gmail dot com
I've executed:
echo $MySQL_DatabaseName . PHP_EOL;
After calling to connect() and it prints out "foo", the value of
$MySQL_DatabaseName.
------------------------------------------------------------------------
[2008-01-29 00:33:45] [EMAIL PROTECTED]
What do you get if you do:
echo $MySQL_DatabaseName . PHP_EOL;
right after your connect() ?
------------------------------------------------------------------------
[2008-01-27 22:40:01] joel dot a dot villarreal at gmail dot com
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
------------------------------------------------------------------------
[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.
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=43947&edit=1
