php password_verify() hash 和 pass 不匹配 [英] php password_verify() hash and pass won't match

问题描述

我将我的密码存储在我用 password_hash() 散列的数据库中，并且我试图在登录时使用 password_verify() 验证密码.出于某种原因，password_verify() 一直返回 false.

I store my passwords in my database hashed with password_hash(), and I am trying to verify the passwords on login with password_verify(). For some reason password_verify() keeps returning false.

我阅读了关于这个函数的文档，它说要确保函数中使用的哈希值在单引号 ' ' 之间，否则它会读取哈希值，因为 $ 是三个变量，所以我试过了像这样写 $valid '$valid'.但这没有用.

I read the documentation on this function and it said to make sure that the hash used in the function is between single quotes ' ' otherwise it will read the hash like it is three variables because of the $'s, so i tried writing $valid like this '$valid'. But that didn't work.

当我回显 $valid 时，输出为 $2y$10$zzZCN7UlukvY2skb3ELVp.4y3Oc7NJTEsFyqdstqYxT

When I echo $valid the output is $2y$10$zzZCN7UlukvY2skb3ELVp.4y3Oc7NJTEsFyqdstqYxT

当我回显 $check 时，输出是 123，这是用于创建帐户的密码.

When I echo $check the output is 123, which is the password used to create the account.

这是我的 login.php 的一部分，这就是我觉得有问题的地方.

This is the part of my login.php, and this is where I feel the problem is.

$emailLogin = mysqli_real_escape_string($con, $_POST['emailLogin']);

$passLogin = mysqli_real_escape_string($con, $_POST['passLogin']);

$query = "SELECT `pass` FROM `user` WHERE `email`='$emailLogin'";

$result = mysqli_query($con, $query);

$row = mysqli_fetch_array($result);

$pass = $row['pass']; 

$key = VUP($passLogin, $pass);

这是我的verify.php的一部分

This is part of my verify.php

function VUP($check, $valid){

if (password_verify($check, $valid)) {
$key = 1;

} else {
echo 'Invalid password.';
$key = 0;
die();
}
return $key;

}

也是verify.php的一部分

Also part of verify.php

function SHP($password){

$hash = password_hash('$password', PASSWORD_BCRYPT);

return $hash;

}

任何建议都会非常有帮助.

Any advice would be very helpful.

推荐答案

当我回显 $valid 时，输出为 $2y$10$zzZCN7UlukvY2skb3ELVp.4y3Oc7NJTEsFyqdstqYxT"

$2y$10$zzZCN7UlukvY2skb3ELVp.4y3Oc7NJTEsFyqdstqYxT 哈希长度只有 50 并且无效/太短，正如我所说，MySQL 将静默失败；错误报告/检查在这里没有帮助.

$2y$10$zzZCN7UlukvY2skb3ELVp.4y3Oc7NJTEsFyqdstqYxT the hash is only 50 in length and is invalid/too short and as I said, MySQL will fail silently; error reporting/checking would not have helped here.

密码的列长度应该是 60(建议 255)，所以它最初没有正确存储.

The password's column length should be 60 (255 is suggested), so it wasn't stored correctly originally.

您需要清除密码列/或表，增加列的长度，然后重新开始.

You will need to clear your password column/or table, increase your column's length, and start over again.

参考:

• http://php.net/manual/en/function.password-hash.php

因此，建议将结果存储在可以扩展到 60 个字符以上的数据库列中(255 个字符是一个不错的选择)."

<小时>

您还可以将查询修改为:

---

You can also modify your query to read as:

$con = new mysqli("xxx", "xxx", "xxx", "xxx");
if ($con->connect_error) {
    die('Connect Error (' . $con->connect_errno . ') '
            . $con->connect_error);
}

$query = "SELECT `pass` FROM `user` WHERE `email`='$emailLogin'";
$result = $con->query($query);

// error checking on the query
if (!$result) {
    echo "<p>There was an error in query: $query</p>";
    echo $con->error;
}

$row_hash = $result->fetch_array();
if (password_verify($passLogin, $row_hash['pass'])) {
    echo "Success!";
}

<小时>

从我留给 OP 的评论中添加:

Adding from a comment I left to the OP:

您的验证功能需要与您的数据库建立连接，这就是我觉得这里发生的事情(变量范围).因此，您需要使用 global $con; 或将连接(变量)传递给您的函数(在大多数情况下更好).

Your verify function needs to have a connection made to your database, that is what I feel is happening here (variable scope). So you'll need to either use global $con; or pass the connection (variable) to your function (which is better in most cases).

我不知道您是否正在为该函数执行包含"，如果是，那么这就是另一个问题.

I don't know if you're doing an "include" for the function, and if so, then that's what the other problem is.

即:function VUP($con, $check, $valid){ 或 function VUP($check, $valid){ global $con; - 两者都试试.使用 $result = mysqli_query($con, $query) 或 die(mysqli_error($con)); 而不是你现在拥有的那个.

I.e.: function VUP($con, $check, $valid){ or function VUP($check, $valid){ global $con; - Try both. Use $result = mysqli_query($con, $query) or die(mysqli_error($con)); instead of the one you have now.

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