Cod*_*key 6 php mysql git openshift
我对openshift和这些git,ssh编程很新.到目前为止我开发了这么多.安装PHP 5.3,MySQL 5.5,phpMyAdmin 4.0墨盒
这是我在php中的dbcon文件.基本上我需要将我的php应用程序与openshift mysql db连接.这就是我所做的.
<?php
// Database Connection Setting
$dbhost = "127.0.0.1"; // Host name
$dbport = "3308"; // Host port
$dbusername = "user"; // Mysql username
$dbpassword = "pass"; // Mysql password
$db_name = "mf"; // Database name
$mysqlCon = mysqli_connect($dbhost, $dbusername, $dbpassword, "", $dbport) or die("Error: " . mysqli_error($mysqlCon));
mysqli_select_db($mysqlCon, $db_name) or die("Error: " . mysqli_error($mysqlCon));
?>
Run Code Online (Sandbox Code Playgroud)
这给了我一个关于openshift的错误,但适用于其他php应用程序.我的错误解释没有得到任何错误:{...空格...}.
你能救我一下吗?
Cod*_*key 15
非常感谢你的努力.之前没有抛出视觉错误.但是我这样做是为了让它发挥作用.
**
**
define('DB_HOST', getenv('OPENSHIFT_MYSQL_DB_HOST'));
define('DB_PORT', getenv('OPENSHIFT_MYSQL_DB_PORT'));
define('DB_USER', getenv('OPENSHIFT_MYSQL_DB_USERNAME'));
define('DB_PASS', getenv('OPENSHIFT_MYSQL_DB_PASSWORD'));
define('DB_NAME', getenv('OPENSHIFT_GEAR_NAME'));
$dbhost = constant("DB_HOST"); // Host name
$dbport = constant("DB_PORT"); // Host port
$dbusername = constant("DB_USER"); // MySQL username
$dbpassword = constant("DB_PASS"); // MySQL password
$db_name = constant("DB_NAME"); // Database name
Run Code Online (Sandbox Code Playgroud)
**
**
$dbhost = getenv('OPENSHIFT_MYSQL_DB_HOST'); // Host name
$dbport = getenv('OPENSHIFT_MYSQL_DB_PORT'); // Host port
$dbusername = getenv('OPENSHIFT_MYSQL_DB_USERNAME'); // MySQL username
$dbpassword = getenv('OPENSHIFT_MYSQL_DB_PASSWORD'); // MySQL password
$db_name = getenv('OPENSHIFT_GEAR_NAME'); // Database name
Run Code Online (Sandbox Code Playgroud)