javascript - php 网站验证不起作用？

Question

我在尝试检查这两个输入 uname 和 passwod 时遇到了麻烦。我可以让一个单独工作，但是当我尝试将它们都传递回来时，我不断收到错误。

<?php 
// define variables and set to empty values
$usernameErr="";
$passwordErr="";
$username= "";
$password="";

if ($_SERVER["REQUEST_METHOD"] == "POST")
{

    if (empty($_POST["uname"]))
        {
         $usernameErr = " username is required <br>";
         print $usernameErr;
        }

    else
        {
        $username = checkUserData($username);
        }   


    if (empty($_POST["passwd"]))
        {
         $passwordErr = "  password is required <br>";
         print $passwordErr;
        }

    else
        {
        $password = checkUserData($password);
        }   
    }

问题就在这里。我正在向下检查数据以阻止攻击。我尝试了多种方法将它们连接在一起，但一切都失败了。

function checkUserData($username)
{
    $username = htmlspecialchars($username);
    $username = trim($username);
    $username = stripslashes($username);
    return $username;
}

function checkUserData($password)
{
    $password = htmlspecialchars($password);
    $password = trim($password);
    $password = stripslashes($password);
    return $password;
}

我打印只是为了检查它是否正常工作。

print ("welcome " .checkUserData($_POST["uname"]));
print ("welcome " .checkUserData($_POST["passwd"]));
?>

任何帮助都会很棒。

Answer 1

两个函数都在做同样的事情，所以概括它们

function SanitizeData($var)
{
    $var= htmlspecialchars($var);
    $var= trim($var);
    $var= stripslashes($var);
    return $var;
}

现在在您的检查过程中调用

$username = SanitizeData($_POST["uname"]);

或者

$password = SanitizeData($_POST["passwd"]);

Although this sanitization is unnecessary at best and destructive at worst if you are going to use these fields in a query, it would be better to use a parameterized query and the PDO database extension.