php变量没有收到价值

Question

HTML代码，我将其添加到form.html文件中。

<select name="find">
        <option value = "a">I'm a regular customer</option>
        <option value = "b">TV advertising</option>
        <option value = "c">Phone directory</option>
        <option value = "d">Word of mouth</option>

PHP代码。我添加到提交后将跳转到的php页面。

switch($find) {
 case "a" :
    echo "<p>Regular customer.</p>";
    break;
 case "b" :
    echo "<p>TV advertising.</p>";
    break;
 case "c":
    echo "<p>Phone directory.</p>";
    break;
 case "d":
    echo "<p>Customer referred by word of mouth.</p>";
    break;
 default :
    echo "<p>We don't know.</p>";
    break;
}

问题是$find似乎无效。因为php页面总是显示默认情况，当我回显$find时，整个php页面都不会显示，我是php的新手。请帮助!!

Answer 1

您是否尝试添加以下内容？

$find = $_POST['find'];
above php code

Answer 2

试试这个..

forward_list

Answer 3

尝试将文件扩展名从.html更改为.php