php password_hash和password_verify问题不匹配 [英] php password_hash and password_verify issues no match

问题描述

我正在尝试从PHP 5.5中调用一个名为password_hash()的新功能.

无论我做什么，$ hash和$ password都不匹配.

$password = "test";

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



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

解决方案

您的代码存在的问题是，在处理哈希时，您使用的是双引号"而不是单引号'. /p>

分配时:

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

这使php认为您有一个名为$2y的变量，还有一个名为$10的变量，最后是一个名为$fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e的变量.显然不是这样.

我在打开错误报告时注意到该错误:

通知:未定义的变量:fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e

被PHP抛出.

将所有双引号替换为单引号进行修复.

例如

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

将整个哈希处理为文字字符串，而不是带有嵌入式变量的字符串.

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

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";
}

解决方案

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.

When assigning:

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

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.

I noticed when turning on error reporting that the error:

Notice: Undefined variable: fXJEsC0zWAR2tDrmlJgSaecbKyiEOK9GDCRKDReYM8gH2bG2mbO4e

Was being thrown by PHP.

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

E.g

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

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

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