根据动态下拉列表中的选择显示div

Question

全部，

根据数据库结果，我有一个选项是动态的下拉列表。当用户从此下拉列表中进行选择时，我希望div显示与此选择相关的结果（也从数据库中提取，链接到选择）。我有这个混合的html / php / javascript既没有显示任何错误，也没有做我需要的东西。我的问题是，这里有什么问题：

   $result_stat_query = ("SELECT statid, stat_name FROM stats");

    $result_stat = $DB_con->query($result_stat_query);

if(!$result_stat)
            {
                $DB_con->debug_mode('query', 'error', '#Query Failed<br/>'.$DB_con->error);
                return false;
            }       

echo "<select name='stat_id' onchange='filterContent();'>";

while ($row_stat = $result_stat->fetch()) {

              unset($statid, $stat_name);
              $statid = $row_stat['statid'];
              $stat_name = $row_stat['stat_name']; 
              //declared variables now to set the javascript so that the content only appears for each selected drop down item
              ?>
     <script>
function filterContent() {
var user = document.getElementById("stat_id").value;
var contentA = document.getElementById("contentA");
 if(user=="<?php $statid;  ?>") {
    contentA.style.display="block";
}  
</script>
              <?php
              //now grab each id and stat name and display in the drop down
              echo '<option value="'.$statid.'">'.$stat_name.'</option>';
}

echo "</select>";

?> 

<div id="contentA">
   <?php echo "<p>Stat name= $stat_name</p>";  ?>
</div>

修改

我试过这个，但它仍然无效： 在头脑中：

               <script>
function filterContent() {
var user = document.getElementById("stat_id").value;
var contentA = document.getElementById("contentA");
    contentA.style.display="block";

</script>

我的php，html部分看起来像这样：

  $result_stat_query = ("SELECT statid, stat_name, speed, strength, health, temper, armor, alertness, intelligence, emotion FROM character_stats");

    $result_stat = $DB_con->query($result_stat_query);

if(!$result_stat)
            {
                $DB_con->debug_mode('query', 'error', '#Query Failed<br/>'.$DB_con->error);
                return false;
            }       

echo "<select name='stat_id' onchange='filterContent();'>";

while ($row_stat = $result_stat->fetch()) {

              unset($statid, $stat_name);
              $statid = $row_stat['statid'];
              $stat_name = $row_stat['stat_name']; 
              //declared variables now to set the javascript so that the content only appears for each selected drop down item
              ?>

              <?php
              //now grab each id and stat name and display in the drop down
              echo '<option value="'.$statid.'">'.$stat_name.'</option>';
}

echo "</select>";

?> 

    <div id="contentA">
   <?php echo "<p>Stat name= $stat_name</p>";  ?>
</div>