从mysli_fetch_array中获取选定的值并输出(Get selected value from mysli_fetch_array and output)

 我有一个问题是获得一个选定的值并输出它。 示例i从下拉列表中选择1001 。 当我回显它总是返回第一行的值，希望是1002 。 
 
 这是我的代码edit.php 
 
<form id="form" action="test.php" method="post">
    <?php
    echo "<select name=\"Reservation ID\" form=\"form\">";
    while ($row = mysqli_fetch_array($result)) 
    {
        $gg = $row['reserve_id'];
         echo "<option value='" . $gg . "' name=\"reserve_id\">" . $gg . "</option>";
    }
    echo "</select>";
    $_SESSION['reserve'] = $gg;
    ?>
    <input type="submit" name="form" value="Submit">
</form>
 
 这是来自test.php的代码 
 
$y = $_SESSION['reserve'];
if(isset($_POST['form']))
{
  echo $y;
}
 
 下拉值 
 
 预订ID一览 

I have a problem to get a selected value and output it. Example i select 1001 from dropdown. When I echo it always return the value from first row wish is 1002.
 
this my code edit.php
 
<form id="form" action="test.php" method="post">
    <?php
    echo "<select name=\"Reservation ID\" form=\"form\">";
    while ($row = mysqli_fetch_array($result)) 
    {
        $gg = $row['reserve_id'];
         echo "<option value='" . $gg . "' name=\"reserve_id\">" . $gg . "</option>";
    }
    echo "</select>";
    $_SESSION['reserve'] = $gg;
    ?>
    <input type="submit" name="form" value="Submit">
</form>
 
this is code from test.php
 
$y = $_SESSION['reserve'];
if(isset($_POST['form']))
{
  echo $y;
}
 
dropdown value
 
List of reservation ID

原文：https://stackoverflow.com/questions/42605087