Php 拒绝用户访问''@'；本地主机'；（使用密码：否）使用AWS和EC2

当使用ShifteditIDE尝试连接到运行LAMP服务器和mysql服务器的AmazonEC2实例时，我遇到以下错误

我用PHP编写的连接到sql server的代码如下：

<?php
function connect_to_database() {

$link = mysqli_connect("localhost", "root", "test", "Jet");

if (!$link) {
    echo "Error: Unable to connect to MySQL." . PHP_EOL;
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
}

echo "Success: A proper connection to MySQL was made! The my_db database is great." . PHP_EOL;
echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;

mysqli_close($link);

}
?>

输出：成功：正确连接到MySQL！my_db数据库是 伟大的主机信息：本地主机通过UNIX套接字访问被拒绝 用户“”@'localhost'（使用密码：否）

我确实使用了正确的root密码，因为我可以在使用Phpmyadmin时成功登录，我只是因为某些原因无法与PHP建立连接

目前，我有一个安装了LAMP服务器和MySQL服务器的AmazonEC2实例。任何帮助都将不胜感激

---

编辑：我使用的是PHP5.6.17

当你在一个函数/方法中创建一个mysqli实例（通过

newmysqli（…）

或

mysqli\u connect（…）

时，你必须考虑Php。从函数中创建mysqli实例，并让调用方使用和/或分配该实例

<?php
/*
builds and throws an exception from the error/errno properties of a mysqli or mysqli_stmt instance
@param useConnectError true: use connect_error/connect_errno instead
@throws mysqli_sql_exception   always does
*/
function exception_from_mysqli_instance($mysqli_or_stmt, $useConnectError=false) {
    // see http://docs.php.net/instanceof
    if ( !($mysqli_or_stmt instanceof mysqli) && !($mysqli_or_stmt instanceof mysqli_stmt) {
        // see docs.php.net/class.mysqli-sql-exception
        throw new mysqli_sql_exception('invalid argument passed');
    }
    else if ($useConnectError) {
        // ok, we should test $mysqli_or_stmt instanceof mysqli here ....
        throw new mysqli_sql_exception($mysqli_or_stmt->connect_error, $mysqli_or_stmt->connect_errno);
    }
    else {
        throw new mysqli_sql_exception($mysqli_or_stmt->error, $mysqli_or_stmt->errno);
    }

}
/* creates a new database connection and returns the mysqli instance
@throws mysqli_sql_exception   in case of error
@return valid mysqli instance
*/
function connect_to_database() {
    $link = new mysqli("localhost", "root", "test", "Jet");
    // see http://docs.php.net/mysqli.quickstart.connections
    if ( $link->connect_errno) {
        // a concept you might or might not be interested in: exceptions
        // in any case this is better than to just let the script die
        // give the other code components a chance to handle this error  
        exception_from_mysqli_instance($link, true);
    }

    return $link;
}

try { // see http://docs.php.net/language.exceptions
  // assign the return value (the mysqli instance) to a variable and then use that variable
  $mysqli = connect_to_database();

  // see http://docs.php.net/mysqli.quickstart.prepared-statements
  $stmt = $mysqli->prepare(....)
  if ( !$stmt ) {
    exception_from_mysqli_instance($stmt);
  }
  ...
}
catch(Exception $ex) {
  someErrorHandler();
}

---

在函数/方法中创建mysqli实例（通过
新建mysqli（…）
或
mysqli_connect（…）
时，必须考虑php。从函数中创建mysqli实例，并让调用方使用和/或分配该实例

<?php
/*
builds and throws an exception from the error/errno properties of a mysqli or mysqli_stmt instance
@param useConnectError true: use connect_error/connect_errno instead
@throws mysqli_sql_exception   always does
*/
function exception_from_mysqli_instance($mysqli_or_stmt, $useConnectError=false) {
    // see http://docs.php.net/instanceof
    if ( !($mysqli_or_stmt instanceof mysqli) && !($mysqli_or_stmt instanceof mysqli_stmt) {
        // see docs.php.net/class.mysqli-sql-exception
        throw new mysqli_sql_exception('invalid argument passed');
    }
    else if ($useConnectError) {
        // ok, we should test $mysqli_or_stmt instanceof mysqli here ....
        throw new mysqli_sql_exception($mysqli_or_stmt->connect_error, $mysqli_or_stmt->connect_errno);
    }
    else {
        throw new mysqli_sql_exception($mysqli_or_stmt->error, $mysqli_or_stmt->errno);
    }

}
/* creates a new database connection and returns the mysqli instance
@throws mysqli_sql_exception   in case of error
@return valid mysqli instance
*/
function connect_to_database() {
    $link = new mysqli("localhost", "root", "test", "Jet");
    // see http://docs.php.net/mysqli.quickstart.connections
    if ( $link->connect_errno) {
        // a concept you might or might not be interested in: exceptions
        // in any case this is better than to just let the script die
        // give the other code components a chance to handle this error  
        exception_from_mysqli_instance($link, true);
    }

    return $link;
}

try { // see http://docs.php.net/language.exceptions
  // assign the return value (the mysqli instance) to a variable and then use that variable
  $mysqli = connect_to_database();

  // see http://docs.php.net/mysqli.quickstart.prepared-statements
  $stmt = $mysqli->prepare(....)
  if ( !$stmt ) {
    exception_from_mysqli_instance($stmt);
  }
  ...
}
catch(Exception $ex) {
  someErrorHandler();
}

您在那里运行的是哪个php版本？当使用php 5.5或更高版本时，作为mysqli php库。如果您尝试以其他用户身份登录，会发生什么情况？创建一个，如果您还没有其他用户，请尝试。为了安全起见：您可以在
mys之后插入一些输出，如
echo'connection closed'；
qli_关闭（$link）；
行，然后向我们显示输出内容（同时编辑代码段，以便代码和输出同步）？@ShashankShah嗨，我用的是PHP5.6。17@SageArslan您好，我尝试过以各种其他用户的身份登录-当使用php 5.5或更高版本时，我会收到与您在那里运行的php版本相同/类似的错误消息？与mysqli php库相同。如果您尝试以其他用户的身份登录，会发生什么情况？创建一个，如果没有其他用户，请尝试然而，为了安全起见：你能在
mysqli_close（$link）；
行之后插入一些输出，比如
echo'connection closed'；
，然后告诉我们输出是什么（同时编辑代码片段，以便代码和输出同步）？@ShashankShah嗨，我用的是PHP5.6。17@SageArslan您好，我尝试过以其他用户的身份登录-我收到了相同/类似的错误消息