这个问题在这里已有答案:
我有一个代码,但有一个问题。它是一个登录表单,用于检查用户并通过,如果没有问题,它将创建一个会话。在另一个页面中,它将检查会话登录是否与true相反。如果确实如此,它将登录。问题是当用户想要登录时,他/她需要提交登录表单2次,以便他/她可以进入网站。问题是什么?
<form method="post" attribute="post" action="test.php">
<p>Username<br/>
<input type="text" id="user" name="user" class="form-control" required></p>
<p>Password<br/>
<input type="text" id="pass" name="pass" class="form-control" required></p>
<p></p>
<button type="submit" name="sub2" id="sub" value="sub" class="btn btn-default btn-block">Login</button>
</form>
<?PHP
include 'config.php';
$user = $_POST["user"];
$pass=$_POST["pass"];
if (isset($_POST['user']) and isset($_POST['pass']))
{
$conn = new mysqli($servername, $username, $password, $dbname);
$result = $link->query("SELECT user FROM users2 WHERE user = '$user'");
$userpass = $link->query("SELECT pass FROM users2 WHERE user = '$user'");
$row = $userpass->fetch_assoc();
$userpasss = $row["pass"];
if($result->num_rows == 1 and $pass == $userpasss)
{
session_start();
$_SESSION["login"] = true;
$_SESSION["username"] = "$user";
echo "hello";
}
}
?>
<?PHP
session_start();
echo $_SESSION["login"];
echo $_SESSION["username"];
?>
任何帮助赞赏。谢谢你的亲切问候
试试这段代码:
$_SESSION["username"] = $user;
它应该工作