PHP抛出解析错误：语法错误，意外'&引用'；

我制作了一个系统，员工可以在其中查看他们申请的休假状态。我正在尝试根据用户是否登录显示申请的休假状态

下面是一个例子：

因此，登录的用户是John Doe，但出于某种原因，我可以看到其他用户申请的休假。我只想让约翰·多伊看看他自己的申请表

下面是我想到的代码：

<div class="container">
    <div class="page-header">
        <h3>My Leaves</h3>
        <div class="table-responsive">
            <table class="table">
                <tr>
                    <th>Employee Name</th>
                    <th>Phone</th>
                    <th>Email</th>
                    <th>From</th>
                    <th>To</th>
                    <th>Reason</th>
                    <th>Status</th>
                </tr>
                <?php
                include ('database.php');
                $result = $database->prepare ("SELECT leaves.* FROM leaves INNER JOIN employee ON employee.id = leaves.user_id WHERE employee ON employee.id = leaves.user_id WHERE employee.username = $_SESSION["VALID_USER_ID"]");
                $result ->execute();
                for ($count=0; $row_message = $result ->fetch(); $count++){ ?>
                <tr>
                    <td><?php echo $row_message['full_name']; ?></td>
                    <td><?php echo $row_message['phone']; ?></td>
                    <td><?php echo $row_message['email']; ?></td>
                    <td><?php echo $row_message['fromdate']; ?></td>
                    <td><?php echo $row_message['todate']; ?></td>
                    <td><?php echo $row_message['reason']; ?></td>
                    <td><?php echo $row_message['status']; ?></td>
                </tr>
                <?php } ?>
            </table>
            <a href="home"><button type="button" class="btn btn-primary"><i class="glyphicon glyphicon-arrow-left"></i> Back</button></a>
        </div>
    </div>
</div>

我不知道是什么

以下是我的表格模式：

---

您将按以下方式更改查询并尝试

$result = $database->prepare ("SELECT leaves.* FROM leaves INNER JOIN employee ON employee.id = leaves.user_id WHERE employee ON employee.id = leaves.user_id WHERE employee.id = ".$_SESSION["VALID_USER_ID"]);

---

只需将结尾改为

employee.username=“.$\u SESSION[“VALID\u USER\u ID”]；

非常感谢。所有申请的休假详细信息都消失了，但是的，我想这是另一个问题。再次感谢

$result = $database->prepare ("SELECT leaves.* FROM leaves INNER JOIN employee ON employee.id = leaves.user_id WHERE employee ON employee.id = leaves.user_id WHERE employee.id = ".$_SESSION["VALID_USER_ID"]);