好的我正在尝试拉出与登录用户名相对应的特定行数据。在我的页面上我已经开始了我的会话但由于某种原因我无法让代码工作。我不断得到一个“数组到字符串转换为“在线”WHERE username ='$ _SESSION [user]'“);”我做错了什么?如果我设置用户名=设置的用户名它可以工作,但我需要它从会话ID中绘制所以它将根据登录的人显示不同的值。
<?php
require("common.php");
if(empty($_SESSION['user']))
{
header("Location: login.php");
die("Redirecting to login.php");
}
?>
<?php
$con=mysqli_connect("localhost","root","nathan","site");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM users
WHERE username = '".$_SESSION['user']."'");
while($row = mysqli_fetch_array($result))
{
echo $row['username'] . " " . $row['att'];
echo "<br>";
}
?>
here are my other corresponding files
<?php
$username = "root";
$password = "nathan";
$host = "localhost";
$dbname = "site";
$options = array(PDO::MYSQL_ATTR_INIT_COMMAND => 'SET NAMES utf8');
try
{
$db = new PDO("mysql:host={$host};dbname={$dbname};charset=utf8", $username, $password, $options);
}
catch(PDOException $ex)
{
die("Failed to connect to the database: " . $ex->getMessage());
}
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);
if(function_exists('get_magic_quotes_gpc') && get_magic_quotes_gpc())
{
function undo_magic_quotes_gpc(&$array)
{
foreach($array as &$value)
{
if(is_array($value))
{
undo_magic_quotes_gpc($value);
}
else
{
$value = stripslashes($value);
}
}
}
undo_magic_quotes_gpc($_POST);
undo_magic_quotes_gpc($_GET);
undo_magic_quotes_gpc($_COOKIE);
}
header('Content-Type: text/html; charset=utf-8');
session_start();
和我的登录文件
<?php
require("common.php");
$submitted_username = '';
if(!empty($_POST))
{
$query = "
SELECT
id,
username,
password,
salt,
email
FROM users
WHERE
username = :username
";
$query_params = array(
':username' => $_POST['username']
);
try
{
$stmt = $db->prepare($query);
$result = $stmt->execute($query_params);
}
catch(PDOException $ex)
{
die("Failed to run query: " . $ex->getMessage());
}
$login_ok = false;
$row = $stmt->fetch();
if($row)
{
$check_password = hash('sha256', $_POST['password'] . $row['salt']);
for($round = 0; $round < 65536; $round++)
{
$check_password = hash('sha256', $check_password . $row['salt']);
}
if($check_password === $row['password'])
{
$login_ok = true;
}
}
if($login_ok)
{
unset($row['salt']);
unset($row['password']);
$_SESSION['user'] = $row;
header("Location: private.php");
die("Redirecting to: private.php");
}
else
{
print("Login Failed.");
$submitted_username = htmlentities($_POST['username'], ENT_QUOTES, 'UTF-8');
}
}
?>
<h1>Login</h1>
<form action="login.php" method="post">
Username:<br />
<input type="text" name="username" value="<?php echo $submitted_username; ?>" />
<br /><br />
Password:<br />
<input type="password" name="password" value="" />
<br /><br />
<input type="submit" value="Login" />
</form>
<a href="register.php">Register</a>
答案 0 :(得分:0)
您收到此错误的原因很简单..您的$_SESSION['user'];
是不是一个数组..这个错误消息的一个简单示例:
$Array = array (1,2,3,4,5);
echo $Array;
那是你的错误..
所以我的建议没有看到实际的$_SESSION
变量..是执行:
echo "<pre>";
print_r($_SESSION);
echo "</pre>";
这将显示整个会话数组,并告诉你我告诉你的确切内容。
你说的话。
$Username_Values = $_SESSION['user'];
$Username_Values['username']; // Will output the user name
答案 1 :(得分:-1)
也许这会有用吗?
<?php
session_start();
$con=mysqli_connect("localhost","root","xxx","xxxx");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM users
WHERE username = '".$_SESSION['user']."'");
while($row = mysqli_fetch_array($result))
{
echo $row['username'] . " " . $row['att'];
echo "<br>";
}
?>