仅在person_id正确时显示项目 - Only show items when person_id is correct?

Only show items when person_id is correct?

本文关键字：显示项目 person id 仅在 | 更新日期: 2023-09-27

>我目前正在尝试编写一段代码，我想在我的页面中显示数据库的 2 个单元格，但前提是person_id是正确的。

我以前使用过占位符，但似乎找不到在 OOP 中执行此操作的方法？

$connect = new MySQL('localhost','username','password');
$connect->Database('dbname');
$posts = $connect->Fetch('group_table WHERE person_id = :user');
$posts->execute[('user' => $_SESSION['id'])];
if ($posts && mysql_num_rows($posts) > 0) {
 echo "Here is some post data:<BR>";
 while ($record = mysql_fetch_array($posts)) {
  ?>
<div class="row">
<div class="col-lg-6 white">
    <?php echo $record['groupname'];?>
        <?php echo $record['group_omschrijving'];?>
    </div>
</div>

我已经通过声明变量 for 来修复它$person

$posts = $connect->Fetch('group_table WHERE person_id = "$person_id"');
$person_id = $_SESSION['id'];