选择会话id为且语法正确的php函数时出错

晚上好，我正在做大学作业，我在显示仅显示与会话id相关的项目列表时遇到问题

我想我做错了

 $query = $db->prepare("SELECT * FROM tbBiodata  where idfunc  like '%$currentUser['id']%' ");

此变量的id为：$currentUser['id']

我在页面的开头使用以下代码来输入变量：

 $currentUser = $user->getUser();

和展览清单

<?php
require_once "db.php";
require_once "User.php";

// Create a user object
$user = new User($db);
// If not logged in
if(!$user->isLoggedIn()){
    header("location: login.php"); //Redirect to the login page
}
// Retrieve current user data
$currentUser = $user->getUser();

//===============

// Create a prepared statement to retrieve all data from tbBiodata
$query = $db->prepare("SELECT * FROM tbBiodata  where idfunc  like '%$currentUser['id']%' ");


// Run the SQL command
$query->execute();
// Take all data and enter variable $ data
$data = $query->fetchAll();
?>

---

当变量包含带引号的数组索引时，需要将其放入
{}
中：

$query = $db->prepare("SELECT * FROM tbBiodata  where idfunc  like '%{$currentUser['id']}%' ");

但是，最好使用参数而不是替换变量

$query = $db->prepare("SELECT * FROM tbBiodata  where idfunc  like CONCAT('%', :id, '%') ");
$query->bindParam(":id", $currentUser['id']);

 <h3>hi<font color="red"><?php echo $currentUser['nama'] ?></font>, <a href="logout.php">Logout</a></h3>

$query = $db->prepare("SELECT * FROM tbBiodata  where idfunc  like '%{$currentUser['id']}%' ");

$query = $db->prepare("SELECT * FROM tbBiodata  where idfunc  like CONCAT('%', :id, '%') ");
$query->bindParam(":id", $currentUser['id']);