首先是代码：
事务处理.php

<tbody>

    <?php
    include 'database_connection.php';
    $sql = "SELECT `id`, `date`, `name`, `phone`, `subtotal`, `status` FROM `sales`";
    $result = $conn->query($sql);

    while ($row = mysqli_fetch_array($result)):
        ?>
        <tr>
            <td><?php echo $row['date']; ?></td>
            <td> <?php echo $row['id'] ?></td>
            <td><?php echo $row['name']; ?></td>
            <td><?php echo $row['phone']; ?></td>
            <td><?php echo $row['subtotal']; ?></td>
            <td><?php echo $row['status']; ?> 
            <td><button type="button" class="btn btn-success btn-xs" id="<?php echo $row['id'];?>" onclick='view(this.id)'>View / Print</button>
        </tr>
    <?php endwhile; ?>
    </tbody>
</table>

<script>

<script>
function view(id){

    $.ajax({
        url: "invoice_view",
        type:"post",
        data:{ val : id },

        success: function(result){

            window.open ("invoice_view.php","mywindow");

    }
    });
}
</script>

发票视图.php

<?php
    include 'database_connection.php';
    $data=$_POST['val'];
    $user_id=$data;
    $sql = "SELECT `name`, `phone`, `town`, `date`, `item1`, `item2`, `item3`, `item4`, `item5`, `item6`, `item7`, `item8`, `q1`, `q2`, `q3`, `q4`, `q5`, `q6`, `q7`, `q8`, `taxper`, `amt1`, `amt2`, `amt3`, `amt4`, `amt5`, `amt6`, `amt7`, `amt8`, `taxtotal`, `subtotal`, `status` FROM `sales` WHERE id=$user_id";
    $result = $conn->query($sql);

    while ($row = mysqli_fetch_array($result)):
        ?>

因此，我尝试将该特定事务的id传递到invoice_view.php页面，以便它加载该特定事务。
transactions.php页面的屏幕截图是：http://oi65.tinypic.com/2ngh24l.jpg
请帮我找出哪里做错了？invoice\u视图甚至没有收到我从transactions页面发送的值。