如何使用php(托管在dotCloud上)连接到远程mysql数据库
我无法连接到我的数据库驻留在dotCloud。我想:如何使用php(托管在dotCloud上)连接到远程mysql数据库
$mysqli = new mysqli($db_host, $db_user, $db_password, $db_name);
和
$mysqli = mysqli_connect($db_host, $db_user, $db_password, $db_name);
和
$mysqli = new mysqli($remote_server, $db_user, $db_password, $db_name);
和
$mysqli = mysqli_connect($remote_server, $db_user, $db_password, $db_name);
,但连接失败,我也得到“错误324(净值:: ERR_EMPTY_RESPONSE):服务器关闭没有sendin的连接g任何数据。“
我找回动态上面下面的mysqli的脚本变量:
$env = json_decode(file_get_contents("/home/dotcloud/environment.json"));
$db_user = $env->DOTCLOUD_DB_MYSQL_LOGIN;
$db_password = $env->DOTCLOUD_DB_MYSQL_PASSWORD;
$db_host = $env->DOTCLOUD_DB_MYSQL_HOST;
$db_port = $env->DOTCLOUD_DB_MYSQL_PORT;
$remote_server = '$db_host:$db_port';
//I also define $db_name here
$db_name = 'mydbname';
我也有mysqli的脚本下面的下面的代码:
if (mysqli_connect_error()) {
die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
$result["status"] = "failed";
$result["message"] = "Failed to connect to database.";
echo json_encode($result);
exit;
} else {
// Successfully connected!
$stmt = $mysqli->stmt_init();
echo "<p>Successfully connected!!</p>";
}
我在做什么错?
您的代码中有几个错误。
1.您$ remote_server的变量是使用单引号
$remote_server = '$db_host:$db_port';
这意味着$ remote_server的不会扩大$ DB_HOST和$ DB_PORT变量。你应该使用双引号。如果你使用了这个变量,它就不适合你。
$remote_server = "$db_host:$db_port";
更多信息,请参见本页面: http://www.php.net/manual/en/language.types.string.php#language.types.string.syntax.single
2.您不使用连接时,MySQL的端口,这是需要在dotCloud因为它不标准端口上运行mysql 3306
您的代码:
$mysqli = new mysqli($db_host, $db_user, $db_password, $db_name);
正确的代码,使用变量,你上面已经宣布:
$mysqli = new mysqli($db_host, $db_user, $db_password, $db_name, $db_port);
更多信息可以在这里找到:http://www.php.net/manual/en/mysqli.quickstart.connections.php
对于一个完整的例子,它看起来像这样。
$env = json_decode(file_get_contents("/home/dotcloud/environment.json"));
$db_user = $env->DOTCLOUD_DB_MYSQL_LOGIN;
$db_password = $env->DOTCLOUD_DB_MYSQL_PASSWORD;
$db_host = $env->DOTCLOUD_DB_MYSQL_HOST;
$db_port = $env->DOTCLOUD_DB_MYSQL_PORT;
//I also define $db_name here
$db_name = 'mydbname';
$mysqli = new mysqli($db_host, $db_user, $db_password, $db_name, $db_port);
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
echo $mysqli->host_info . "\n";
try like this: I think it will help you.
$connect = mysql_connect($mysql_host, $mysql_user, $mysql_password);
if (!$connect) {
die('Could not connect: ' . mysql_error());
}
//echo 'Connected successfully';
$db = mysql_select_db($mysql_database,$connect);
if (!$db) echo"'Could not select database";
谢谢您的建议。我试过了,它给了我这个错误(和页面没有加载):错误324(net :: ERR_EMPTY_RESPONSE):服务器关闭连接而不发送任何数据。 – Naomi 2013-03-02 01:37:18
Outch! **请不要举例使用mysql_ *。此扩展从PHP 5.5.0开始已弃用。** – 2013-03-02 01:59:07
您是否可以使用MySQL客户端(SQLYog,MySQL Workbench等)连接到MySQL主机? – juanchopx2 2013-03-02 02:56:25
您的$ db_name变量在哪里? – Stepo 2013-03-02 01:07:09
@Stepo,我编辑了上面的代码以显示我定义$ db_name的位置。 – Naomi 2013-03-02 03:26:47