php password_hash 和 password_verify 问题不匹配

问题描述

我正在尝试 PHP 5.5 中的一个名为 password_hash() 的新函数.

I am trying out a new function from PHP 5.5 called password_hash().

无论我做什么，$hash 和 $password 都不会匹配.

No matter what i do the $hash and the $password wont match.

$password = "test";

$hash = "$2y$10$fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e";



if (password_verify($password, $hash)) {
    echo "Success";
}
else {
    echo "Error";
}

推荐答案

你的代码的问题是你使用双引号 " 而不是单引号 ' 处理哈希时.

The problem with your code is that you are using the double quotation marks " instead of the single quotation marks ' when dealing with your hash.

分配时:

$hash = "$2y$10$fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e";

这让 php 认为你有一个名为 $2y 的变量和另一个名为 $10 的变量，最后是第三个名为 $fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e 的变量.显然不是这样的.

It's making php think you have a variable called $2y and another one called $10 and finally a third one called $fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e. Which obviously isn't the case.

我在打开错误报告时注意到错误:

I noticed when turning on error reporting that the error:

注意:未定义变量:fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e

被 PHP 抛出.

用单引号替换你所有的双引号来修复.

Replace all your double quote marks with single quote marks to fix.

例如

$hash = '$2y$10$fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e';

将整个哈希视为文字字符串，而不是带有嵌入变量的字符串.

Treats the whole hash as a literal string instead of a string with embedded variables.

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