Using php I need to have 2 different connections to the database - php

Question

I have a need to copy some rows from a database to a different database. I am experiencing difficulty. I have found several methods except none of the seem to work. The php version I am using is 5.4.
Both connections are in the same server, however everything else is different
This is the php code that I have found and it doesnt seem to work at all, I am unable to select from the first database
// Create connection
$wpdb = mysql_connect($servername, $username, $password);
// Check connection
if ($wpdb->connect_error) {
die("Connection failed: " . $wpdb->connect_error);
}
echo "Connected local successfully\n";
//$starttime = date("h:i:sa");
$mydb = mysql_connect('localhost','dbname','dbpassword', true);
// Check connection
if ($mydb->connect_error) {
die("Connection failed: " . $mydb->connect_error);
}
echo "Connected to Integrity successfully\n";
mysql_select_db($database, $wpdb);
mysql_select_db('wordpress_0', $mydb);

Answer 1

you can try with PDO.that provide a common interface to talk with many different databases.
$pdo = new PDO('mysql:host=example.com;dbname=database', 'user',
'password');

Answer 2

I refuse to offer support for mysql_ syntax, so I'll offer the upgraded version.
Almost entirely copied from the php manual... http://php.net/manual/en/mysqli.select-db.php
Code:
/* attempt and check connection including first database selection "test" */
if (!$mysqli = new mysqli("localhost", "root", "", "test")) {
// never show the actual error to the public
echo "<div>Database Connection Error: " , $conn->connect_error , "</div>";
exit();
}
/* return name of current default database */
if (!$result = $mysqli->query("SELECT DATABASE()")) {
// never show the actual error to the public
echo "<div>Syntax Error: " , $conn->error , "</div>";
} else {
echo "<div>Default database is: " , $result->fetch_row()[0] , "</div>";
$result->close();
}
/* change db to "mysql" db */
$mysqli->select_db("mysql");
/* return name of current default database */
if (!$result = $mysqli->query("SELECT DATABASE()")) {
// never show the actual error to the public
echo "<div>Syntax Error: " , $conn->error , "</div>";
} else {
echo "<div>Default database is: " , $result->fetch_row()[0] , "</div>";
$result->close();
}
$mysqli->close();
Output:
Default database is: test
Default database is: mysql