PHP抛出解析错误：语法错误，意外'“' [重复]

Question

我创建了一个系统，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来显示已申请的休假状态。

这是一个例子：

所以登录的用户是 John Doe，但由于某种原因，我可以看到其他用户提出的其他申请。我只希望 John Doe 看到他自己的应用的叶子。

这是我想出的代码：

<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>

不幸的是，我收到了Parse error: syntax error, unexpected '"'

在下面一行：

$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"]");

我不确定它是什么。

这是我的表架构：

Answer 1

您将如下更改查询并尝试。

$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"]);