PHP echo ID和Pop Up表单

Question

我有一个弹出窗体，当用户点击＆＃34;更新＆＃34;时，应该有一个弹出窗体，用户应该可以更新但弹出窗口不会出现，因为＆＃34; a HREF＆＃34;有回声＆＃34;在id值。见下文：echo $ orderId

<a href="#id=<?php echo $orderId ?>" id="btn">Update</a>

但是，如果我删除<?php echo $orderId ?>部分，例如<a href="#id" id="btn">Update</a>，则弹出窗口有效，但回显错误ID。

以下是表格

<a href="#x" class="overlay" id="id=<?php echo $orderId; ?>"></a>
<div class="popup">
<form method = "post" action="">
    <div>
        <label for="Order ID">Order ID:</label><?php echo $orderId ?>
    </div><br>
    <div>
        <label for="Order Date">Order Date:</label><?php echo $orderDate ?>
    </div><br>
    <div>
        <label for="Order Total">Order Total:</label><?php echo $totalAmount ?>
    </div><br>
    <div>
        <label for="Preparation Status">Preparation Status:</label>
            <select name="preparationStatus"/>
                <option value="<?php echo $preparationStatus; ?>">Default(<?php echo $preparationStatus; ?>)</option>
                <option value="Pending">Pending</option>
                <option value="Prepared">Prepared</option>
            </select>
    </div><br>
    <div>
        <label for="Payment Status">Payment Status:</label>
            <select name="paymentStatus">
                <option value="<?php echo $paymentStatus;?>">Default(<?php echo $paymentStatus;?>)</option>
                <option value="Pending">Pending</option>
                <option value="Paid">Paid</option>
            </select>
    </div><br>
        <input type="submit" id="btn" name="submit" value="Save" />
        <a class="close" href="#close"></a>
</form>

它只回显id="id=1"

如何显示弹出窗体？