如何从html中的下拉框中传递SESSION值

Question

我有一个包含项目列表的下拉值代码，我想将我选择的值发布到另一个名为checkout.php的页面中

这是我的代码：

<?php
session_start();

if ($_POST['submit'])
    $_SESSION['movies'] = $_POST['movie'];
?>

<form name="booking" action="checkout.php" onsubmit="return validateBooking()" method="post" target="_blank">
    <p class="selection"> Movies</p></br>
    <select name="movie">
        <option value="empty">Please Select a Movie</option>
        <option value="AC">Dunkirk</option>
        <option value="RC">Annabelle</option>
        <option value="CH">The Dark Tower</option>
        <option value="AF">Logan Lucky</option>
    </select>
</form>

Answer 1

像这样使用，先给action=""，这样表单就会发布在当前页面上，然后将数据存储在会话中并重定向到checkout.php。现在在结帐页面，您可以从会话中获取数据

<?php
    session_start();
    if ($_POST['movie']){
        $_SESSION['movies'] = $_POST['movie'];
        header('Location: checkout.php');
    }
     ?>

    <form name="booking" action="" onsubmit="return validateBooking()" method="post">
        <p class="selection"> Movies </p></br>
        <select name="movie">
            <option value="empty">Please Select a Movie</option>
            <option value="AC">Dunkirk</option>
            <option value="RC">Annabelle</option>
            <option value="CH">The Dark Tower</option>
            <option value="AF">Logan Lucky</option>
        </select>
   </form>

你可以在checkout.php上写同样的php代码当你给action='checkout.php'时

<form name="booking" action="checkout.php" onsubmit="return validateBooking()" method="post">

checkout.php 包含

<?php
    session_start();
    if ($_POST['movie']){
        $_SESSION['movies'] = $_POST['movie'];
    }
?>