Php 当重定向回起始页时，如何显示在下拉菜单中选择的原始名称？

我想返回到我的原始索引页面，让下拉菜单显示最初选择的名称。然后，我想选择另一个学生并再次执行任务

目前，我能够：

从下拉菜单中选择学生并将所选学生提交到另一页

更新学生的特征并重定向回原始页面

我可以在下拉菜单中显示学生姓名，但当我选择一个新学生时，我选择的正上方的学生将被选中

第1页代码，带下拉菜单

 <div>
    <?php
     if (session_status() == PHP_SESSION_NONE  || session_id() == '') {
        session_start();
        $pid = $_SESSION['stu_name_id_select'];
        echo $pid;
       }
     ?>
    <form name = "test" method="POST" action = "">
    <?php
    $select_student= $db->prepare("SELECT stu_name_id FROM students WHERE active = 'Yes' order by stu_name_id");    //sql select query
    $select_student->execute();
    echo "<select name='stu_name_id_select' onChange ='this.form.submit()'><option>Select</option>";   
    while ($result1=$select_student->fetch(PDO::FETCH_ASSOC)){
       echo "<option value = '".$pid."'";
      //echo "<option value = '".$result1['stu_name_id']."'";
      if (isset($_POST['stu_name_id_select']) && $_POST['stu_name_id_select'] == $result1['stu_name_id'])  echo 'selected="selected"';
         echo ">".$result1['stu_name_id']."</option>";  
         $pid = $resule1['stu_name_id'];
        }
    echo "</select>";       
   ?>
 <input type="submit" value="Add/Edit Records"  onclick="test.action='edit_records.php'; return true;" />
 <input type="submit" value="Add/Edit Accommodations"  onclick="test.action='accommodations.php'; return true;" />
 </form>
 </div>

---

您将在循环的末尾而不是开始处重新分配

$pid

，因此每个学生实际上都有上一个学生的pid

只需移动

$pid=$result1['stu_name_id']

作为

while

循环启动后的第一条指令

while ($result1=$select_student->fetch(PDO::FETCH_ASSOC)){
  $pid = $resule1['stu_name_id'];
  echo "<option value = '".$pid."'";
  if (isset($_POST['stu_name_id_select']) && $_POST['stu_name_id_select'] == $result1['stu_name_id'])  echo 'selected="selected"';
  echo ">".$result1['stu_name_id']."</option>";  
}

while（$result1=$select\u student->fetch（PDO:：fetch\u ASSOC））{
$pid=$result1['stu_name_id']；
回显“$result1['stu_name_id'”；
}

我通过更改此代码解决了此问题

while ($result1=$select_student->fetch(PDO::FETCH_ASSOC)){
  echo "<option value = '".$pid."'";
  if (isset($_POST['stu_name_id_select']) && $_POST['stu_name_id_select'] == $result1['stu_name_id'])  echo 'selected="selected"';
  echo ">".$result1['stu_name_id']."</option>";  
  $pid = $result1['stu_name_id'];
}

while（$result1=$select\u student->fetch（PDO:：fetch\u ASSOC））{
回显“$result1['stu_name_id'”；
$pid=$result1['stu_name_id']；
}

到

foreach（$result1作为$row）{
如果（isset（$\u会话['stu\u name\u id\u select']）和&$\u会话['stu\u name\u id\u select']=$row['stu\u name\u id']））
{
回显“$”会话['stu\u name\u id\u select']；
$\u POST['stu\u name\u id\u select']=$\u会话['stu\u name\u id\u select']；
取消设置（$_会话['stu\u name\u id\u select']；）
否则{
回显“$row['stu_name_id']。”；}
} 

必须首先将

会话启动

功能发送到浏览器，否则它将无法正常工作。你的不是第一件事…@Kinglish回声$pid显示的是正确的名称，所以我认为这部分工作正常。如果我删除这一行，$pid=$result1['stu_name_id']；下拉菜单显示正确的名称，但由于$pid未更新，因此我无法选择新学生。当我移动$pid=$Result1['stu_name_id']；我可以从下拉窗口中选择学生的姓名，在我的选择上方选择学生的问题不会发生，但当我返回到索引页时，所选学生的姓名不再显示在下拉窗口中，现在显示select。

while ($result1=$select_student->fetch(PDO::FETCH_ASSOC)){
  echo "<option value = '".$pid."'";
  if (isset($_POST['stu_name_id_select']) && $_POST['stu_name_id_select'] == $result1['stu_name_id'])  echo 'selected="selected"';
  echo ">".$result1['stu_name_id']."</option>";  
  $pid = $result1['stu_name_id'];
}

  foreach ($result1 as $row){
     if(isset($_SESSION['stu_name_id_select']) && $_SESSION['stu_name_id_select'] == $row['stu_name_id'])
            {
            echo "<option selected = 'selected'  value=\"".$_SESSION['stu_name_id_select']."\">".$_SESSION['stu_name_id_select']."</option>";
            $_POST['stu_name_id_select'] = $_SESSION['stu_name_id_select'];
            unset($_SESSION['stu_name_id_select']);}
           else{    
             echo "<option value = '".$row['stu_name_id']."'";
             if (isset($_POST['stu_name_id_select']) && $_POST['stu_name_id_select'] == $row['stu_name_id']) echo 'selected="selected"';
             echo ">".$row['stu_name_id']."</option>";}
    }