试图从通过为每个显示的表中回显一列

Question

我正在使用这张表并试图在警报中呼应“branchid” 

有两个表：

订单地址.php

<?php

  session_start();

  require_once('orders_address.vc.php');

?>

这是我对每张桌子的片段，“branchid”和分配按钮只是这里的问题

                    <td>

                      <a href="order_address.vc.php<?php echo '?branchid='.$rowAddress['branchid']; ?>">

                        <input type="submit" class="btn button-color-blue font-color-white full_width" name="assign" value="ASSIGN">

                      </a>

                    </td>

                    <td class="table-text-center">

                      <?php

                          echo($rowAddress['branchid']);

                      ?>

                    </td>

订单地址.vc.php

下面是点击按钮时的代码

  if (isset($_POST['assign']) && $_POST['assign'] == 'ASSIGN')

  {

    $branchid = $_GET ['branchid'];

    echo "<script type='text/javascript'>alert('$branchid');</script>";

    }

目前我得到一个未定义的索引并且警告框是空的。$_GET ['branchid'] 似乎没有检索到我想要的列。

感谢您的任何帮助。

Answer 1

<td>&nbsp; &nbsp; <?php&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; $branchid = isset($rowAddress['branchid']) ? $rowAddress['branchid'] : 0;&nbsp;&nbsp; &nbsp; ?>&nbsp; &nbsp; <form method="POST" action="order_address.vc.php">&nbsp; &nbsp; &nbsp; &nbsp; <input type="hidden" name="branchid" value="<?php echo $branchid ?>" />&nbsp; &nbsp; &nbsp; &nbsp; <input type="hidden" name="assign" value="ASSIGN" />&nbsp; &nbsp; &nbsp; &nbsp; <input type="submit" value="ASSIGN" class="btn button-color-blue font-color-white full_width" />&nbsp; &nbsp; </form></td><td class="table-text-center">&nbsp; <?php echo $branchid; ?></td>orders_address.vc.php:if(isset($_POST['assign']) && $_POST['assign'] === 'ASSIGN') {&nbsp; &nbsp; &nbsp;&nbsp; &nbsp; echo '<script type="text/javascript">';&nbsp; &nbsp; echo 'alert('.$_POST['branchid'].')';&nbsp; &nbsp; echo '</script>';}

Answer 2

我不知道为什么你有一个围绕锚标记的表单提交按钮。当提交标签被点击时，老实说，我不知道哪个优先，锚标签或表单提交。假设您将表格包装在 HTML 表单中并发出 POST 请求。我建议添加branchid作为提交按钮的值。前任：&nbsp;<input type="submit" class="..." name="assign" value="<?=$rowAddress['branchid']?>">在接收端，您可以branchid从$_POST['assign']if (! empty($_POST['assign'])){&nbsp; &nbsp; $branchid = $_POST['assign'];&nbsp; &nbsp; echo "<script type='text/javascript'>alert('$branchid');</script>";}

Answer 3

因为你$_GET['branchid']在POST处理if&nbsp;(isset($_POST['assign'])&nbsp;&&&nbsp;$_POST['assign']&nbsp;==&nbsp;'ASSIGN')&nbsp;...但是您像往常一样发送GEThref 链接，因此条件POST永远不会为真。

Answer 4

if (isset($_POST['assign']) && $_POST['assign'] == 'ASSIGN'){$branchid = $_GET ['branchid'];echo "<script type='text/javascript'>alert('$branchid');</script>";}HTML 应如下所示：<form action="order_address.vc.php" method="POST">&nbsp; &nbsp; &nbsp; &nbsp; <input type="hidden" name="branchid" value="<?php echo $branchid ?>" />&nbsp; &nbsp; &nbsp; &nbsp; <input type="submit" name="assign" value="ASSIGN" /></form>