简体   繁体   English

无法将用户输入到password_verify()

[英]Can't get user input in to password_verify()

here is the login code, I'm trying to get password verify to work but It doesn't want to. 这是登录代码,我正在尝试让密码验证工作,但它不想。 It seems like this count($sql->fetchAll()) > 0 is the problem or that I am calling the wrong variable to the password_verify() . 看起来这个count($sql->fetchAll()) > 0是问题,或者我正在调用password_verify()的错误变量。

    $signup = isset($_POST['signup'] ) ? $_POST['signup'] : false;
$submit = isset( $_POST['submit'] ) ? $_POST['submit'] : false;
$username = isset( $_POST['username'] ) ? $_POST['username'] : false;
$password = isset( $_POST['password'] ) ? $_POST['password'] : false;

if( $submit && $username && $password ) {

  $sql = $DB_con->prepare( "SELECT * FROM users WHERE user_name=:user_name AND user_pass=:user_pass" );
  $sql->bindParam( ':user_name', $username, PDO::PARAM_STR );
  $sql->bindParam( ':user_pass', $password, PDO::PARAM_STR );
  $check_user=$sql->fetch(PDO::FETCH_ASSOC);
  $success = $sql->execute();
  $verify = password_verify($password, check_user['user_pass']);
    // Successfully logged in!
  if($success && count($sql->fetchAll()) > 0 && $verify) {

    $_SESSION['logged_in'] = true;  
    // Unset errors
    unset($_SESSION['error']);
  }
  else {
    // display an error
    $_SESSION['error'] = 'That user doesn\'t exist'; 
  }
}
else {
  //displays error
  $_SESSION['error'] = 'Please enter all fields';
}
if($signup) {
  header('Location: signup.php');
}
exit;

Here is the signup code 这是注册码

function register($uname,$umail,$upass)
    {

        $DB_con = null;

try
{
     $DB_con = new PDO("mysql:host=$host;dbname=$dbname", "$username" ,"$password");
     $DB_con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $e)
{
     echo $e->getMessage();
}   



       try
       {
           $new_password = password_hash($upass, PASSWORD_DEFAULT);

           $stmt = $DB_con->prepare("INSERT INTO users(user_name,user_email,user_pass) 
                                                       VALUES(:uname, :umail, :upass)");

           $stmt->bindparam(":uname", $uname);
           $stmt->bindparam(":umail", $umail);
           $stmt->bindparam(":upass", $new_password);            
           $stmt->execute(); 
           return $stmt; 

       }
       catch(PDOException $e)
       {
           echo $e->getMessage();
       }    

    }

There are several things that needs to be cleaned up with this piece of code. 使用这段代码需要清理几件事。

Let's start with your register() function. 让我们从register()函数开始。 There's a clear syntax-error here (where you make the connection), the ""$password" . Then that function don't know what $dbname , $username or $password are (for the connection), these variables aren't in the scope of the function (it can see it). Usually it's better to pass the connection object as a parameter, not create a new object for each time you call the register() function. Modified for that and cleaned a bit, it would look like this 这里有一个明确的语法错误(你在哪里建立连接), ""$password" 。然后那个函数不知道$dbname$username$password是什么(对于连接),这些变量不是在函数的范围内 (它可以看到它)。通常最好将连接对象作为参数传递,而不是每次调用register()函数时都创建一个新对象。为此修改并清理了一下,它看起来像这样

function register($DB_con, $uname ,$umail, $upass) {
    try {
        $new_password = password_hash($upass, PASSWORD_DEFAULT);

        $stmt = $DB_con->prepare("INSERT INTO users (user_name, user_email, user_pass) 
                                                   VALUES (:uname, :umail, :upass)");

        $stmt->bindparam(":uname", $uname);
        $stmt->bindparam(":umail", $umail);
        $stmt->bindparam(":upass", $new_password);            
        return $stmt->execute(); 

    } catch(PDOException $e) {
        echo $e->getMessage();
    }
}

and be used like 并被用作

register($DB_con, 'Nader', 'example@example.com', 'pa$$w0rd'); // True or false

Then onto your sign-in. 然后进入你的登录。 This is a little messy, and you're doing a few things wrong. 这有点乱,你做错了。 What I've found so far is... 到目前为止我发现的是......

  • Fetching after execution 执行获取
  • Selecting WHERE the unhashed password is equal to the hashed one (won't return any rows) 选择WHERE未使用的密码等于哈希密码(不会返回任何行)
  • count($sql->fetchAll()) > 0 will be zero, because you already fetched once, and there will only be one user => this means one row. count($sql->fetchAll()) > 0将为零,因为您已经获取了一次,并且只有一个用户=>这意味着一行。 Fetching one means there are no more to fetch, so fetchAll() is empty! 获取一个意味着没有更多要获取,所以fetchAll()是空的!
  • You don't need to do any more checks than if ($check_user=$sql->fetch()) to check if there are rows 您不需要再执行任何检查if ($check_user=$sql->fetch())来检查是否有行

Corrected and accounted for the points above, your code would look something like this 更正并计算了上述各点,您的代码看起来像这样

$signup = isset($_POST['signup']) ? $_POST['signup'] : false;
$submit = isset($_POST['submit']) ? $_POST['submit'] : false;
$username = isset($_POST['username']) ? $_POST['username'] : false;
$password = isset($_POST['password']) ? $_POST['password'] : false;

if ($submit && $username && $password) {
    $sql = $DB_con->prepare("SELECT * FROM users WHERE user_name=:user_name");
    $sql->bindParam(':user_name', $username, PDO::PARAM_STR);
    $sql->execute();

    if ($check_user=$sql->fetch(PDO::FETCH_ASSOC)) {
        if (password_verify($password, $check_user['user_pass'])) {
            $_SESSION['logged_in'] = true;  
            // Unset errors
            unset($_SESSION['error']);
        } else {
            // Invalid password
        }
    } else {
        // No user with that username
    }
} else {
    //displays error
    $_SESSION['error'] = 'Please enter all fields';
}

if ($signup) {
    header('Location: signup.php');
}
exit;

This assumes unique usernames in the DB 这假定数据库中有唯一的用户名

There might be more issues which I can't see, which is why you should enable error-reporting by adding 可能会有更多我无法看到的问题,这就是您应该通过添加启用错误报告的原因

<?php
error_reoprting(E_ALL);
ini_set("display_errors", 1);

at the top of your file (errors shouldn't be displayed like that in a live environment). 在文件的顶部(错误不应该像在实时环境中那样显示)。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM