跨页面传递会话用户 ID

Question

我需要传递会话用户名（他们已登录），这是一封电子邮件！我需要将此传递到单独的页面并将其输出到表格中以表示提交的评论$_SESSION['name'] = $_POST['name']; - 刷新时发送页面以登录 $name = ['name'] - 将页面发送回登录<!-- logged in user information --><?php  if (isset($_SESSION['email'])) : ?><p>Welcome you are logged in as: <strong><?php echo $_SESSION['email']; ?></strong></p>“电子邮件”需要从“index.php”上方的代码传递到“reviews.php”下方的代码<p><input name="product_id" value="<?php echo "$var" ?>" readonly> <!-- get value from previous page--><input name="track_name" value="<?php echo "$var_value" ?>" readonly> <!-- get value from previous page--><input name="track_name" value="<?php echo "EMAIL_HERE" ?>" readonly> <!-- get value from previous page--><!-- get value from previous page--><input type="Submit" name="Submit" value="Submit"></p>由于这是一项作业，我只能使用 PHP MYSQL HTML CSS我希望用户名（电子邮件）在表中显示为 $var 和 $var_value 是，然后它们应该全部以表格形式并排打印出来使用此代码更新我现在设法获取变量值，但无法将其插入数据库$email = $_SESSION['email'];$sql = "INSERT INTO reviews (rating, review, track_name, product_id, email) values('$rate', '$text', '$track', '$artist', '$email')";``"只读>```所以更新是我现在如何将它插入到我的数据库中？

Answer 1

最后感谢大家的帮助

$email = $_SESSION['email']; 从会话中获取电子邮件

$email = (isset($_POST['email']) ? $_POST['email'] : null); 删除索引错误

<input name="email" value="<?php echo "$email" ?>" readonly> <!-- get value from previous page--> 显示值

$review_query = mysqli_query($result,"SELECT rating, review, email FROM reviews WHERE track_name = '$var_value' AND product_id = '$var'"); 从数据库中获取

<td class='col-4 col-s-4' name='email'><?php echo $email ?></td> 输出它的值

Answer 2

您是否尝试在 review.php 页面中使用 $_SESSION['email'] 变量？我希望你也使用 CSRF 代币