PHP 请求未验证密码是否正确，然后继续打开 HTML 按钮

Question

我不知道在这里做什么。PHP post请求（request.php）只是在单击按钮（correct.html）后继续下一页，即使我没有在输入中键入任何内容。我希望它验证密码是否正确，如果是这样=继续更正.html，否则提醒用户。PHP 代码：<?php $pass = $_POST['password'];if ($pass == "password"){   include "correct.html";}else{   echo "Password incorrect";}?>代码：<div class="input-container">    <input class="input-field" type="password" placeholder="Your passphrase" /><br>    <i class="fa fa-user icon"></i></div><br><div id=continue>    <form action="request.php" method="post">     <button class="button" name="password" value="password" style="vertical-align:middle"><span>Confirm</span></button>    </form></div>

Answer 1

您应该首先为输入类型密码或按钮添加名称属性值，如下所示

<div id=continue>

        <form action="request.php" method="post">

            <input class="input-field" type="password" name="password" placeholder="Your passphrase" />

            <button class="button" name="submit" value="submit" style="vertical-align:middle"><span>Confirm</span></button>

        </form>

</div>

之后你的php代码应该是这样的

    <?php

    if(isset($_POST['submit']))

    {

    $pass = $_POST['password'];

    if ($pass == "password")

    {

       include "correct.html";

    }

    else

    {

       echo "Password incorrect";

    }

    }

?>

Answer 2

这是你必须具备的：form

<div id=continue>

    <form action="request.php" method="post">

        <!-- Note that input and button are under the SAME <form> -->

        <input class="input-field" type="password" name="password" placeholder="Your passphrase" />

        <!-- Note `name` attributes of `button` and `input`  -->

        <button class="button" name="button" value="submit" style="vertical-align:middle"><span>Confirm</span></button>

    </form>

</div>

在服务器上：

<?php

// for debugging purposes, remove when you want

print_r($_POST);

// as 'password' is now NAME of INPUT, `$pass` stores the value from the INPUT

$pass = $_POST['password'];

if ($pass == "password")

{

   include "correct.html";

}

else

{

   echo "Password incorrect";

}

Answer 3

您应该阅读有关 PHP 中的表单提交的信息。在此特定实例中，您的表单未作为按钮需要提交属性提交提交，即 .<button type="submit"

此外，要调试 post 数据，可以使用 。var_dump($_POST); die();