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>&nbsp; &nbsp; &nbsp; &nbsp; <form action="request.php" method="post">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <input class="input-field" type="password" name="password" placeholder="Your passphrase" />&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <button class="button" name="submit" value="submit" style="vertical-align:middle"><span>Confirm</span></button>&nbsp; &nbsp; &nbsp; &nbsp; </form></div>之后你的php代码应该是这样的&nbsp; &nbsp; <?php&nbsp; &nbsp; if(isset($_POST['submit']))&nbsp; &nbsp; {&nbsp; &nbsp; $pass = $_POST['password'];&nbsp; &nbsp; if ($pass == "password")&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp;include "correct.html";&nbsp; &nbsp; }&nbsp; &nbsp; else&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp;echo "Password incorrect";&nbsp; &nbsp; }&nbsp; &nbsp; }?>

Answer 2

这是你必须具备的：form<div id=continue>&nbsp; &nbsp; <form action="request.php" method="post">&nbsp; &nbsp; &nbsp; &nbsp; <!-- Note that input and button are under the SAME <form> -->&nbsp; &nbsp; &nbsp; &nbsp; <input class="input-field" type="password" name="password" placeholder="Your passphrase" />&nbsp; &nbsp; &nbsp; &nbsp; <!-- Note `name` attributes of `button` and `input`&nbsp; -->&nbsp; &nbsp; &nbsp; &nbsp; <button class="button" name="button" value="submit" style="vertical-align:middle"><span>Confirm</span></button>&nbsp; &nbsp; </form></div>在服务器上：<?php// for debugging purposes, remove when you wantprint_r($_POST);// as 'password' is now NAME of INPUT, `$pass` stores the value from the INPUT$pass = $_POST['password'];if ($pass == "password"){&nbsp; &nbsp;include "correct.html";}else{&nbsp; &nbsp;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();