我不知道在這里做什么。PHP post請求（request.php）只是在單擊按鈕（correct.html）后繼續下一頁，即使我沒有在輸入中鍵入任何內容。我希望它驗證密碼是否正確，如果是這樣=繼續更正.html，否則提醒用戶。PHP 代碼：<?php $pass = $_POST['password'];if ($pass == "password"){ include "correct.html";}else{ echo "Password incorrect";}?>代碼：<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>
3 回答
瀟湘沐
TA貢獻1816條經驗 獲得超6個贊
您應該首先為輸入類型密碼或按鈕添加名稱屬性值，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代碼應該是這樣的
<?php
if(isset($_POST['submit']))
{
$pass = $_POST['password'];
if ($pass == "password")
{
include "correct.html";
}
else
{
echo "Password incorrect";
}
}
?>
小怪獸愛吃肉
TA貢獻1852條經驗 獲得超1個贊
這是你必須具備的：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>
在服務器上：
<?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";
}
慕尼黑的夜晚無繁華
TA貢獻1864條經驗 獲得超6個贊
您應該閱讀有關 PHP 中的表單提交的信息。在此特定實例中，您的表單未作為按鈕需要提交屬性提交提交，即 .<button type="submit"
此外，要調試 post 數據，可以使用 。var_dump($_POST); die();
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