Question

我建立了一个系统，员工可以查看他们的申请休假状态。我正在尝试根据用户是否登录来显示应用的休假状态。

以下是一个例子：

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的其他应用假。我只希望John Doe看到他自己的应用的叶子。

这是我提出的代码：

<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>

不幸的是我得到了Parse error: syntax error, unexpected '"'

在以下一行：

$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。

以下是我的表格模式：

Answer 1

您将更改以下查询并尝试。

$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"]);

PHP抛出Parse错误：语法错误，意外'''

1 个答案: