试图理解PHP password_hash

Question

I'm trying to learn more about PHP security best practices, and I came across password_hash and the password_compat project by Anthony Ferrara. I think I understand how to implement it, but in testing, I came across a curious behavior that contradicts my novice understanding of password hashing.

If I save the result of the password_hash function into a MySQL database user record and then retrieve that hash for verification using password_verify, it works as expected. However, if I do the exact same thing, but rather than pull from the database, I just hard-code the password hash via a copy/paste from the database, the password_verify function fails.

The code is below:

// Get the Username and password hash from the MySQL database.  GetPassTestuser routine returns an array where
//   position[0][0] is the username and position[0][1] is the password hash.
$arrUser = GetPassTestuser("mike24");
echo("User Name: ".$arrUser[0][0]."<br/>");
echo("Password hash: ".$arrUser[0][1]."<br/>");

// Run password_verify with the password hash collected from the database.  Compare it with the string "mytest"
//  (This returns true in my tests).
if (password_verify("mytest",$arrUser[0][1])){
    echo("Password verified");
} else {
    echo("Password invalid");
}
echo("<hr>Now On to our second test...<br/>");
// Now run password_verify with a string representation directly copied/pasted from the database.  This is 
//   being compared with "mytest", which in my mind should return a true value.  But it doesn't and this test
//   fails.  Not sure why.
if (password_verify("mytest","$2y$10$S33h20qxHndErOoxJL.sceQtBQXtSWrHieBtFv59jwVwJuGeWwKgm")){  // String shown here is the same as value contained in $arrUser[0][1]
    echo("2nd Test Password verified");
} else {
    echo("2nd test Password invalid");
}

While this isn't something I'd do in real code, I'm just trying to understand what the difference is. Why does it work OK when I use the string variable that supposedly contains the exact same hash, but doesn't work when it's hard coded?

Thanks!

Answer 1

FROM PHP DOC

To specify a literal single quote, escape it with a backslash (). To specify a literal backslash, double it (\). All other instances of backslash will be treated as a literal backslash: this means that the other escape sequences you might be used to, such as or , will be output literally as specified rather than having any special meaning.

If a dollar sign ($) is encountered, the parser will greedily take as many tokens as possible to form a valid variable name. Enclose the variable name in curly braces to explicitly specify the end of the name.

Use single quotes

Replace

"$2y$10$S33h20qxHndErOoxJL.sceQtBQXtSWrHieBtFv59jwVwJuGeWwKgm"

With

'$2y$10$S33h20qxHndErOoxJL.sceQtBQXtSWrHieBtFv59jwVwJuGeWwKgm'

Or

"\$2y\$10\$S33h20qxHndErOoxJL.sceQtBQXtSWrHieBtFv59jwVwJuGeWwKgm"