Question

我在网站上有这个表格。我无法从select标签中获取值，以便通过php上的变量收集。无论选择哪个选项，它都会在PHP中显示为空白。

其他输入值可以通过它的php变量准确收集。

我相信我已经检查了足够的错别字，未封闭的标签或缺少字符。它会是某种关于＆lt;的规则吗？选择＆gt;我不知道的标签？

HTML

<form method="post" action="contact.php" name="contactform" id="contactform" autocomplete="off">
    <fieldset>
        <label id="sendtolabel"><span>Send To:</span></label>
        <select name="select" id="select">
            <option value="1">Option 1</option>
            <option value="2">Option 2</option>
            <option value="3">Option 3</option>
            <option value="4">Option 4</option> 
        </select>
        <label for="name" accesskey="U"><span class="required">Name</span></label>
        <input name="name" type="text" id="name" size="30" title="name" />
        <label for="email" accesskey="E"><span class="required">Email</span></label>
        <input name="email" type="text" id="email" size="30" title="email" />
        <label for="comments" accesskey="C"><span class="required">Comments</span></label>
        <textarea name="comments" cols="40" rows="3" id="comments" title="comments"></textarea>
        <input type="submit" class="submit" id="submit" value="Submit" />
     </fieldset>
  </form>

PHP

$select = $_POST['select']; 
$name = $_POST['name']; 
$email = $_POST['email']; 
$comments = $_POST['comments']

Answer 1

它在这里工作......你可以试试这个：

<?php
   if ( isset( $_POST['select'] ) ) {
       var_dump( $_POST['select'] );
       $select = $_POST['select']; 
       $name = $_POST['name']; 
       $email = $_POST['email']; 
       $comments = $_POST['comments'];
       echo "THE SELECT VAR's Value is: __".$select;
   }
?>
<form method="post" action="" name="contactform" id="contactform" autocomplete="off">
   <fieldset>
        <label id="sendtolabel"><span>Send To:</span></label>
        <select name="select" id="select">
            <option value="1">Option 1</option>
            <option value="2">Option 2</option>
            <option value="3">Option 3</option>
            <option value="4">Option 4</option> 
        </select>
        <label for="name" accesskey="U"><span class="required">Name</span></label>
        <input name="name" type="text" id="name" size="30" title="name" />
        <label for="email" accesskey="E"><span class="required">Email</span></label>
        <input name="email" type="text" id="email" size="30" title="email" />
        <label for="comments" accesskey="C"><span class="required">Comments</span></label>
        <textarea name="comments" cols="40" rows="3" id="comments" title="comments"></textarea>
        <input type="submit" class="submit" id="submit" value="Submit" />
     </fieldset>
   </form>

注意：您可以在phpfiddle中测试上面的代码。

1）转到http://phpfiddle.org/
  2）然后将代码粘贴到 Editor 中   按运行[F9]

点击此处测试小提琴中的代码：http://phpfiddle.org/main/code/tuc-6jm

Answer 2

当我遇到这样的问题时，我只打印所有$ _POST全局以查看已返回的内容

print_r($_POST);

而且没有..你的代码是正确的

<html>
    <head></head>
    <body>
       <form method="post" action="test.php" name="contactform" id="contactform" autocomplete="off">
          <fieldset>
          <label id="sendtolabel"><span>Send To:</span></label>
          <select name="select" id="select">
              <option value="1">Option 1</option>
              <option value="2">Option 2</option>
              <option value="3">Option 3</option>
              <option value="4">Option 4</option> 
          </select>
          <label for="name" accesskey="U"><span class="required">Name</span></label>
          <input name="name" type="text" id="name" size="30" title="name" />
          <label for="email" accesskey="E"><span class="required">Email</span></label>
          <input name="email" type="text" id="email" size="30" title="email" />
          <label for="comments" accesskey="C"><span class="required">Comments</span></label>
          <textarea name="comments" cols="40" rows="3" id="comments" title="comments"></textarea>
          <input type="submit" class="submit" id="submit" value="Submit" />
         </fieldset>
     </form>
<?php 

   print_r( $_POST );
   echo $_POST['select'];
   $select = $_POST['select']; 
   $name = $_POST['name']; 
   $email = $_POST['email']; 
   $comments = $_POST['comments'];
   echo $select;
?>
    </body>
</html>

选择返回它必须返回的内容...（1,2,3,4）

Answer 3

对我来说很好。

试试这种方式。将此代码放在contact.php之上，以测试是否所有值都正确检索。

<?php
    if ( $_POST ) {
       echo '<pre>';
       print_r( $_POST );
       die;
    }
?>

请参阅下面的输出：http://d.pr/i/dgN3

Select变量中的值在PHP变量上显示为空白

3 个答案: