Php my sql
| From: | I.T Manager ISOSS.NET | Date: | Mon, 21 Mar 2011 15:42:37 +0000 |
| Subject: | Php my sql | ||
| Groups: | php.webmaster | ||
| Request: | Send a blank email to php-webmaster+get-10667@lists.php.net to get a copy of this message | ||
Dear Sir
I have a website (www.123.com) with online library books to download and the
website is connected with PHP MYSQL at the same server where the website is
hosted and the PHP MYSQL database is also hosted at that server to upload
books i want to make some changes according to this :
My other website(www.456.com) which is also hosted at other hosting company
"Host Gator" i want to inter link my PHP MYSQL of (www.123.com) to the PHP
MYSQL of (www.456.com) .
For example if i upload the books at (www.123.com) by using PHP MYSQL then i
wish that the books directly uploads to the hosting space of (www.456.com)
and also i need to update all data at PHP MYSQL of (www.456.com) but i do
not want to use the hosting space of (www.123.com)
Scenario :
1- I upload the book named "Development in Statistics" at (www.123.com) via
php my sql from its cpanle of website but i want that the php mysql should
upload it directly to other website hosting space which is (www.456.com)
and also update the entry in php mysql of (www.456.com)
2- I will import the database backup of (www.123.com) to the php mysql of
(www.456.com) so after that is it possible to auto updation in the php mysql
of (www.456.com) in the targeted database of (www.123.com) ?
Please guide me in this data base string that how i can make changes in it
to connect with other host .
I want that the php mysql will upload books to the other host (www.456.com)
and i will make a new folder at (www.456.com) in which the books will upload
and i want to use php my sql of (www.123.com) and i didnot want to use
hosting space of (www.123.com).
Please guide me
My connection link is :
<?php
$user="123_user";
$password="12345678";
$host="localhost";
$database="123_database";
$conn=mysql_connect($host,$user,$password);
mysql_select_db($database);
$op=$select;
Regards