Javascript 无法接收代码

嘿，伙计们，很抱歉问这个问题，但是我被难住了，我希望当单选按钮被点击时，代码能够转到带有动作的页面，就像愚蠢的单选按钮到dumby.php

 <form name = "quoted" method="get">
 <input id = "poster" type="text" name="poster" required="required" placeholder = "Credited Individual.">     <br>
 <textarea class = "actual_quote" name = "actual_quote" required="required" placeholder = "Write the question here!"></textarea><br><br><br>
 <div class = "checkboxes" required="required">
<h3 style = "margin-top:-20px;">Please select one catagory that the quote falls into.</h3>
  <label for="x"><input type="radio" name="x" value="stupid" id = "x" checked="checked" action = "stupid.php" />    <span>stupid</span></label><br>
    <label for="x"><input type="radio" name="x" value="stupider" id = "x" action = "stupider.php" /> <span>stupider</span>    </label><br>
    <label for="x"><input type="radio" name="x" value="stupidest" id = "x" action = "stupidest.php" />    <span>stupidest</span></label>
  </div>
  <input id = "submit1" type="submit"><br>
 </form>

请选择报价所属的一个类别。
愚蠢的

愚蠢者

愚蠢的

这是每个页面上的php，愚蠢的，愚蠢的，愚蠢的，愚蠢的，php错误是说实际的引用和海报没有被识别索引，帮助

    <div class="wrapper">
    <div class="submissions">
    <div class="logo-logo"><h2>Question.</h2>
    <div class="checkboxes"><?= !empty($_GET['x']) ? $_GET['x'] : '' ?>
    </div>

    </div>

 <div class="top-submit"><?php echo '&#8220;' . (!empty($_GET['actual_quote']) ?  $_GET['actual_quote'] : '') . '&#8221;'; $actual_quote = $_GET['actual_quote'];?>
 </div>
 <div class="poster"><?php echo "-" .  (!empty($_GET['poster']) ? $_GET['poster'] :''); $poster =  $_GET['poster'];?>
 <div class = "like">
 <a href = "javascript:countClicksLike();" class = "btn btn-large" style = "color:green;">Like</a>
 <p id = "like" style = "color:green;">0</p>
 </div>
 <div class = "dislike">
 <a href = "javascript:countClicks();" class = "btn btn-large" style = "float:right; color:red;">Dislike</a>
 <p id = "dis" style = "color:red;">0</p>
 </div>
  </div>
 <?php 
 "INSERT INTO submissions(top-submit, poster)
 VALUES ($actual_quote, $poster)";
?>
</div>
</div>

问题。
0
0

---

很抱歉问一下，我已经试了一个小时了，我的想法都没有了。欢迎提出任何建议。

尝试从

action=“stupidweb3.php”

删除到

action=“stupidweb3.php”

同时检查单选按钮代码（保留唯一id并删除操作）

---

设置变量并运行您的查询

很抱歉，请问其中是否有错误？是的。您有一个不需要的HTMl标记（表单）

<form name = "quoted"  action = "stupidweb3.php" method="get">
 <input id = "poster" type="text" name="poster" required="required" placeholder = "Credited Individual.">     <br>
 <textarea class = "actual_quote" name = "actual_quote" required="required" placeholder = "Write the question here!"></textarea><br><br><br>
 <div class = "checkboxes" required="required">
<h3 style = "margin-top:-20px;">Please select one catagory that the quote falls into.</h3>
  <label for="x"><input type="radio" name="x" id="x" value="stupid" />    <span>stupid</span></label><br>
    <label for="x"><input type="radio" name="x" id="x1" value="stupider" /> <span>stupider</span>    </label><br>
    <label for="x"><input type="radio" name="x" id="x2" value="stupidest"  />    <span>stupidest</span></label>
  </div>
  <input id = "submit1" type="submit"><br>
 </form>

<div class="wrapper">
    <div class="submissions">
    <div class="logo-logo"><h2>Question.</h2>
    <div class="checkboxes"><?php echo (!empty($_GET['x']) ? $_GET['x'] : ''); ?>
    </div>

    </div>

 <div class="top-submit"><?php echo '&#8220;' . (!empty($_GET['actual_quote']) ?  $_GET['actual_quote'] : '') . '&#8221;'; $actual_quote = $_GET['actual_quote'];?>
 </div>
 <div class="poster"><?php echo "-" .  (!empty($_GET['poster']) ? $_GET['poster'] :''); $poster =  $_GET['poster'];?>
 <div class = "like">
 <a href = "javascript:countClicksLike();" class = "btn btn-large" style = "color:green;">Like</a>
 <p id = "like" style = "color:green;">0</p>
 </div>
 <div class = "dislike">
 <a href = "javascript:countClicks();" class = "btn btn-large" style = "float:right; color:red;">Dislike</a>
 <p id = "dis" style = "color:red;">0</p>
 </div>
  </div>

</div>
</div>

<?php 
$sql = "INSERT INTO submissions(top-submit, poster)
 VALUES ('$actual_quote', '$poster')";
?>