检查密码强度 - Uncaught ArgumentCountError: Too little arguments to function

Question

如何检查我的 PHP 代码的密码强度？它显示此错误消息：

致命错误：未捕获的 ArgumentCountError：函数 strongpass::check() 的参数太少，1 在第 8 行的 C:\newxampp\htdocs\fyp\index.php 中传递，而在 C:\newxampp\htdocs 中恰好有 2 \fyp\strongpass.php:7 堆栈跟踪：#0 C:\newxampp\htdocs\fyp\index.php(8): strongpass->check('asdgasdfsadfasd') #1 {main} throw in C:\newxampp\ htdocs\fyp\strongpass.php 在第 7 行

class strongpass{
public function check($password,$user){
$denum_pass = strtr($password,'5301!','seoll');
$word_file = '/usr/share/dict/words'; // here is my dictionary
$lc_user = strtolower($user);
$lc_pass = strtolower($password);
$response = "OK";

//password that can't be same as username or reversed username
else if (($user == $password) || ($password == strrev($user))||
(denum_pass == $user) || ($denum_pass == strrev($user))){
$response = " Password cannot be same with username.";}

//password that doesn't contain any word in dictionary
else if (is_readable($word_file)){
if ($fh = fopen($word_file,'r')){
$found = false;
while (! ($found || feof($fh))) { 
$word = preg_quote(trim(strtolower(fgets($fh,1024))),'/');
if (preg_match("/$word/",$lc_pass) || preg_match("/$word/",$denum_pass)){
$found = true;}}
fclose($fh);
if ($found){
$response = 'Password is based on a dictionary word.';}}}
return $response; } } 

类是这样使用的：

$user = ""; if(isset($_POST["password"])){ 
$password = $_POST["password"]; 
include_once("strongpass.php"); 
$strongpass = new strongpass(); 
$response = $strongpass->check($password);
if($response != "OK"){ 
    $status = $response; 
} else { 
    $status = "Password is strong so parsing can continue here."; 
    
}

Answer 1

strongpass->check('asdgasdfsadfasd') 看起来您只是将 $password 传递给 check 方法，而没有传递 $user。

Answer 2

您定义了需要 $password 和 $user 参数的方法：

public function check($password,$user){

但你只传$password:

$response = $strongpass->check($password);

所以也传递$user：

$response = $strongpass->check($password, $user);

但是，我不知道它会如何工作：

$password = "";
$status = ""; 
$user = "";