"mysqli_select_db" asks for a database name (string) and you gave that function an integer ($connect).
The first parameter should be "$connect" and the second parameter "$database":
PHP код:
<?php
////\\\\\\\\\\\\\\\\\\\\\\EDIT The Below Variables\\\\\\\\\\\\\\\\\
////\\\\\\\\\\\\\\\\\\\ONLY EDIT WITHIN THE QUOTATION MARKS ("")\\\
//
$servername = "n10.ultra-h.com"; //Server name (usually "localhost")
$database = "port_5930"; //Database name (often username_databasename)
$username = "port_5930"; //User name for MySQL Database
$password = "*******"; //Password for MySQL User name
//
//// \\\\\\\\\\\\\\\\\\\\\\\DO NOT EDIT BELOW\\\\\\\\\\\\\\\\\\\\\\\
//
$connect = mysqli_connect($servername,$username,$password);
mysqli_select_db($connect, $database) or die(mysqli_error($connect));
?>
You could also combine it directly with "mysqli_connect":
PHP код:
<?php
////\\\\\\\\\\\\\\\\\\\\\\EDIT The Below Variables\\\\\\\\\\\\\\\\\
////\\\\\\\\\\\\\\\\\\\ONLY EDIT WITHIN THE QUOTATION MARKS ("")\\\
//
$servername = "n10.ultra-h.com"; //Server name (usually "localhost")
$database = "port_5930"; //Database name (often username_databasename)
$username = "port_5930"; //User name for MySQL Database
$password = "*******"; //Password for MySQL User name
//
//// \\\\\\\\\\\\\\\\\\\\\\\DO NOT EDIT BELOW\\\\\\\\\\\\\\\\\\\\\\\
//
$connect = mysqli_connect($servername,$username,$password,$database);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
