Question

我请求您提供如何通过URL将变量发送到其他PHP站点的输入：

＆＃13;

  <form action="deny.php" method="get">
  <div align="left"></div>
  <p><span class="style13" style="height: 23px;">
    <select name="deny" class="style28" style="width: 320px">
      <option value="Select" selected="selected">Select</option>
      <option value="Price">Too expensive</option>
      <option value="Agency">Other Agency</option>
    </select>
    </span></p>
  <p><span class="style13" style="height: 23px;">
  <input type="hidden" name="id" value=<? echo $id; ?> />
  </span> </p>
  <p>
  <? echo '<td><a href="deny.php?submit='.$id.'&deny='.$_GET['deny'].'">Send Feedback</a></td>'; ?>
  </p>
</form>

＆＃13;

$id是正确的，但$deny为空

我甚至尝试使用$deny (instead of $_GET['deny']) and $_POST[deny] - 但$deny始终为空。（可以在链接中控制）

感谢您的建议！

BR，

的Stefan

Answer 1

用以下代码替换您的代码：

<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<form action="deny.php" method="get">
  <div align="left"></div>
  <p><span class="style13" style="height: 23px;">
    <select name="deny" class="style28" style="width: 320px">
      <option value="Select" selected="selected">Select</option>
      <option value="Price">Too expensive</option>
      <option value="Agency">Other Agency</option>
    </select>
    </span></p>
  <p><span class="style13" style="height: 23px;">
  <input type="hidden" name="id" value="<?php echo $id; ?>" />
  </span> </p>
  <p>
  <a href="#" id="feedbackLink">Send Feedback</a>
  </p>
</form>
<script type="text/javascript">
    $( document ).ready(function() {
        $('select[name="deny"]').change(function(){
            var link = 'deny.php?submit=<?php echo $id; ?>&deny=';
            $('a#feedbackLink').attr('href', link + $(this).val());
        });
        $('select[name="deny"]').trigger('change');
    });
</script>

您无法在不提交表单的情况下使用$ _GET，$ _POST等变量。

希望这能解决您的问题。

Answer 2

在您将表单提交到当前脚本之前，

$_GET['deny']不会被填充。

请勿尝试使用PHP构建网址。让浏览器做到这一点。只需在表单中放置一个提交按钮，并有适当的输入和操作。那是什么形式。

Answer 3

<form action="deny.php" method="get">
  <div align="left"></div>
  <p><span class="style13" style="height: 23px;">
    <select name="deny" class="style28" style="width: 320px">
      <option value="Select" selected="selected">Select</option>
      <option value="Price">Too expensive</option>
      <option value="Agency">Other Agency</option>
    </select>
    </span></p>
  <p><span class="style13" style="height: 23px;">

  </span> </p>
  <p>
  <input type="submit" value="submit">
  </p>
</form>

Answer 4

<form action="deny.php" method="get">
    <div align="left"></div>
    <p>
        <span class="style13" style="height: 23px;">

            <select name="deny" class="style28" style="width: 320px">
                <option value="Select" selected="selected">Select</option>
                <option value="Price">Too expensive</option>
                <option value="Agency">Other Agency</option>
            </select>
        </span>
    </p>
    <p>
         <span class="style13" style="height: 23px;">
              <input type="hidden" name="id" value="<?php echo $id; ?>" /> 
         </span>
    </p>
    <p>
         <input type="submit" value="Send Feedback" value="submit" />
    </p>
</form>

 <?php
      if (isset($_GET)) {
          var_dump($_GET);
      }
 ?>

Answer 5

<form action="/deny.php?id=<? $id ?>" method="get">

<input name="submit" type="submit" value="Feedback" />

诀窍:-)

谢谢，

的Stefan

Answer 6

尝试这个，它已经过测试。

<form action="your_page.php" method="GET"> <!--you may use POST for security as well -->
      <select name="deny">
        <option value="">Select Option</option>
        <option value="Price">Too expensive</option>
        <option value="Agency">Other Agency</option>
      </select>
      <!-- you msut have $id = something -->
      <input type="hidden" name="id" value="<?php echo $id; ?>" /> 
      <input type="submit" value="Send Feedback" value="submit" />
  </form>

 <?php
      echo "<pre>";print_r(($_GET));
 ?>

PHP：从下拉列表发送变量到URL

6 个答案: