php - 从 mysli_fetch_array 获取选定值并输出

Question

我在获取选定值并将其输出时遇到问题。示例我从下拉列表中选择1001。当我回显时，它总是返回第一行希望的值是 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;
}

dropdown value

List of reservation ID

Answer 1

这当然是duplicate question .

编辑:

循环执行后$gg将指向列表中的最后一个值(本例中为 1002)。我相信您正在尝试访问用户选择的值 <option> <select>的可以通过以下方式完成:

在edit.php中:

<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;//this is not required to get <select> value, but may be relevant to what you are doing otherwise
    ?>
    <input type="submit" name="form" value="Submit">
</form>

在test.php中:

$y = $_SESSION[''];//this is not required to get <select> value, but may be relevant to what you are doing otherwise
if(isset($_POST['form']))
{
    echo $_POST['Reservation_ID'];
}