Question

我想将数据从数据库显示给当前登录的用户。例如，如果我想告诉他什么时候创建了他的账户，我怎么才能告诉他这个价值呢？我的意思是这个价值只是他的账户。不是全部。我试过了

<?php    
$dbhost = '';
$dbuser = '';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
$sql = "Select date from members"

mysql_select_db('');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
      echo "{$row['date']}  <p> ";
} 
?>

但是这会显示该表中的所有日期。我想只显示当前登录用户的日期。谢谢编辑：用户使用电子邮件登录并通过（已散列）这是我的登录功能

function login($email, $password, $mysqli) {
    // Using prepared statements means that SQL injection is not possible. 
    if ($stmt = $mysqli->prepare("SELECT id, username, password, salt 
        FROM members
       WHERE email = ?
        LIMIT 1")) {
        $stmt->bind_param('s', $email);  // Bind "$email" to parameter.
        $stmt->execute();    // Execute the prepared query.
        $stmt->store_result();

        // get variables from result.
        $stmt->bind_result($user_id, $username, $db_password, $salt);
        $stmt->fetch();

        // hash the password with the unique salt.
        $password = hash('sha512', $password . $salt);
        if ($stmt->num_rows == 1) {
            // If the user exists we check if the account is locked
            // from too many login attempts 

            if (checkbrute($user_id, $mysqli) == true) {
                // Account is locked 
                // Send an email to user saying their account is locked
                return false;
            } else {
                // Check if the password in the database matches
                // the password the user submitted.
                if ($db_password == $password) {
                    // Password is correct!
                    // Get the user-agent string of the user.
                    $user_browser = $_SERVER['HTTP_USER_AGENT'];
                    // XSS protection as we might print this value
                    $user_id = preg_replace("/[^0-9]+/", "", $user_id);
                    $_SESSION['user_id'] = $user_id;
                    // XSS protection as we might print this value
                    $username = preg_replace("/[^a-zA-Z0-9_\-]+/", 
                                                                "", 
                                                                $username);
                    $_SESSION['username'] = $username;
            $_SESSION['email'] = $email;
                    $_SESSION['login_string'] = hash('sha512', 
                              $password . $user_browser);
                    // Login successful.
                    return true;
                } else {
                    // Password is not correct
                    // We record this attempt in the database
                    $now = time();
                    $mysqli->query("INSERT INTO login_attempts(user_id, time)
                                    VALUES ('$user_id', '$now')");
                    return false;
                }
            }
        } else {
            // No user exists.
            return false;
        }
    }
}

Answer 1

在SQL末尾使用WHERE column='value'代码（请参阅：http://www.tizag.com/mysqlTutorial/mysqlwhere.php）

示例：

如果您知道数据库中使用的用户ID，请将其存储在变量中（此处称为$usersIDnr）。

然后使用列名称（例如userID）在SQL结尾的以下代码中使用：

WHERE userID='$usersIDnr'

如何仅从当前登录的用户显示数据库中的数据？

1 个答案: