Question

我有两个下拉框，我希望用户从两者中选择值。下拉框中填充了页面加载查询的结果。我希望这些值存储为会话变量，以便在后续查询中使用。

第一个文件= choose.php

<html>
    <form action="choose2.php" method="post">
    <?//db connection stuff taken out from here
        }
// fill classes      
    $result = mysqli_query($con,"SELECT * FROM classes");                                            
    echo "<select id='classlist'>";

    while($row = mysqli_fetch_array($result))
        {
        echo " <option value=" . $row['id'] . ">" . $row['classname'] . "</option>";
        }
    echo "</select>";

    //fill schemes of work
    $result = mysqli_query($con,"SELECT * FROM schemes_of_work");
    echo "<select id='sowlist'>";

    while($row = mysqli_fetch_array($result))
        {
        echo " <option value=" . $row['id'] . ">" . $row['name'] . "</option>";
        }
    echo "</select>";

    mysqli_close($con);
    ?>   
<input type="submit">
</form> 
</html>

第二档= choose2.php

 <?php  session_start(); 
    $_SESSION['classname'] = $_POST["classlist"];
    $_SESSION['sowname'] = $_POST["sowlist"];
    ?>

    <?php echo $_SESSION['classname'];?>
    <?php echo $_SESSION['sowname'];?>

我不能让它工作 - 我在select2上得到一个空页面，并在apache日志中出现以下错误： “未定义的索引：第3行/var/www/assessment/choose2.php中的播种列表，

Answer 1

select标记应具有name属性，以便能够在php POST中使用它。尝试

<select id='classlist' name="classlist">

另一个人也一样。

php选择下拉选择到会话变量

1 个答案: