Question

新手问题。以下代码不会打印任何内容。我不知道为什么请帮助我努力学习。

<HTML>
<head />
<body>


<form action="test.php" method=POST>
username: <input type="text" name="user"><br/>
password: <input type="password" name="pass"><br />
<input type="submit" value="go"><p>
</form

<?php

$user = $_POST['user'];
$pass= $_POST['pass'];

if(($user=="Rascal999")&&($pass=="poo"))echo "Access Granted";
else echo "Access Denied";

?>

</body>
</HTML>

感谢。

Answer 1

您错过了结束标记：

更改

<form action="test.php" method=POST>
username: <input type="text" name="user"><br/>
password: <input type="password" name="pass"><br />
<input type="submit" value="go"><p>
</form

到

<form action="test.php" method=POST>
username: <input type="text" name="user"><br/>
password: <input type="password" name="pass"><br />
<input type="submit" value="go"><p>
</form>

Answer 2

关注Phil Cross的Build和test.php

<?php
if(isset($_POST) && !empty($_POST)){
   $user = $_POST['user'];
   $pass= $_POST['pass'];
  if( $user=="Rascal999" && $pass=="poo" ):
   echo "Access Granted";
  else: 
   echo "Access Denied";
  endif;
}
?>

如果此页面是test.php，请在表单标记中删除操作并将上述代码粘贴到页面的开头

基本的PHP页面不起作用

2 个答案: