我正在研究我的项目。我有登录系统的问题。首先，我想重定向到我想要的页面。其次，我想在导航栏上显示用户名（与登录用户名相同）。
我已经尝试使用“header（）”进行重定向，但它不起作用。它显示“找不到对象”页面。我也尝试了“window.location.assign（）”但它显示的结果与“header（）”相同。
的index.php
<?php
session_start();
?>
<!DOCTYPE html>
<html>
<head>
.
.
.
<title>Online Stock Trading</title>
</head>
<body>
<!-- navbar -->
<nav class="navbar navbar-expand-lg navbar-dark bg-dark">
.
.
.
<form class="form-inline" action="php/login_page.php" method="post">
<input class="form-control mr-sm-2" type="text" placeholder="Username" name="username">
<input class="form-control mr-sm-2" type="password" placeholder="Password" name="password">
<button class="btn btn-success my-2 mr-sm-2" id="login-btn" type="submit">LOGIN</button>
<button type="button" class="btn btn-primary my-2 mr-sm-2" id="sign_up" data-toggle="modal" data-target="#exampleModalLong">SIGN UP</button>
</form>
</nav>
.
.
.
</body>
</html>
login_page.php
<?php
$con = mysqli_connect("localhost","root","","project");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL".mysqli_connect_error();
}
$username = mysqli_real_escape_string($con,$_POST['username']);
$password = mysqli_real_escape_string($con,$_POST['password']);
$result = mysqli_query($con,"SELECT Username, Password FROM trader_data WHERE Username = '$username' AND Password = '$password'")
or die("Failed to query database".mysql_error());
$row = mysqli_fetch_array($result);
if ($row['Username']==$username && $row['Password']==$password)
{
$_SESSION['username'] = $username;
echo "<script>
alert('Login Success!!');
window.location.assign('logged_in.php');
</script>";
}
else
{
echo "<script>
alert('Failed to login!!');
window.location.assign('index.php');
</script>";
}
mysqli_close($con);
?>
logged_in.php
<?php
session_start();
?>
<!DOCTYPE html>
<html>
<head>
.
.
.
<title>Online Stock Trading</title>
</head>
<body>
<!-- navbar -->
<nav class="navbar navbar-expand-lg navbar-dark bg-dark">
.
.
.
<div class="nav-item dropdown">
<a class="nav-link dropdown-toggle" href="#" id="navbarDropdown" role="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false">
<?php echo $_SESSION["username"];?>
</a>
<div class="dropdown-menu" aria-labelledby="navbarDropdown">
<a class="dropdown-item" href="#">My Account</a>
<a class="dropdown-item" href="#">History</a>
<a class="dropdown-item" href="#">Portfolio</a>
<div class="dropdown-divider"></div>
<a class="dropdown-item" href="#">Logout</a>
</div>
</div>
</div>
</div>
</nav>
.
.
.
</body>
</html>
我希望我的网站在登录后会转到“logged_in.php”页面。然后，显示用于登录“index.php”页面的用户名。
您可以尝试使用window.location.href进行重定向。要将值传递到另一个页面，您可以使用会话或发出GET请求。
使用会话：
login_page.php
<?php
session_start();
$_SESSION['username'] = $username;
?>
<script>
window.location.href="logged_in.php";
</script>logged_in.php
<?php
session_start();
$username = $_SESSION['username'];
?>使用Qazxswpoi请求
login_page.php
GETlogged_in.php
<script>
window.location.href="logged_in.php?username=<?php echo $username;?>"
</script>