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如何从 python 生成与 Spring 兼容的密码？

我有一个 spring 应用程序在数据库中创建 scrypt 密码，如下所示：

{scrypt}$e0801$QAqC0fvhY6iJPysiQsFnrcUg205njHo/6o+IDXDn33lxmZOCVBhb4NAqdafhuGmykCxQtMI5xP5zb7MYMUrU3Q==$sBeXCHOm6zQuGdSDKs+HeXnNQGg3bhRidmL+HU/ZTMM=

我正在尝试使用 python 和 passlib 直接使用新密码更新数据库，但无法从以下位置预测 passlib 的正确 setting_kwds $e0801$：

>>> from passlib.hash import scrypt
>>> scrypt.verify('wBkfoBsxj9u3wLOZ', '{scrypt}$e0801$QAqC0fvhY6iJPysiQsFnrcUg205njHo/6o+IDXDn33lxmZOCVBhb4NAqdafhuGmykCxQtMI5xP5zb7MYMUrU3Q==$sBeXCHOm6zQuGdSDKs+HeXnNQGg3bhRidmL+HU/ZTMM=')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/home/gregn/.pyenv/versions/v2.7.16/lib/python2.7/site-packages/passlib/utils/handlers.py", line 789, in verify
    self = cls.from_string(hash, **context)
  File "/home/gregn/.pyenv/versions/v2.7.16/lib/python2.7/site-packages/passlib/handlers/scrypt.py", line 177, in from_string
    return cls(**cls.parse(hash))
  File "/home/gregn/.pyenv/versions/v2.7.16/lib/python2.7/site-packages/passlib/handlers/scrypt.py", line 181, in parse
    ident, suffix = cls._parse_ident(hash)
  File "/home/gregn/.pyenv/versions/v2.7.16/lib/python2.7/site-packages/passlib/utils/handlers.py", line 1207, in _parse_ident
    raise exc.InvalidHashError(cls)
ValueError: not a valid scrypt hash

手动摆弄前缀格式没有奏效：

>>> scrypt.verify('wBkfoBsxj9u3wLOZ', '$scrypt$ln=1,r=8,p=1$QAqC0fvhY6iJPysiQsFnrcUg205njHo/6o+IDXDn33lxmZOCVBhb4NAqdafhuGmykCxQtMI5xP5zb7MYMUrU3Q==$sBeXCHOm6zQuGdSDKs+HeXnNQGg3bhRidmL+HU/ZTMM=')
False

我认为（但不确定）Spring 正在使用默认的 SCryptPasswordEncoder 设置，因为我发现：

public static PasswordEncoder getPasswordEncoder() {
    final String encodingId = "scrypt";
    final Map<String, PasswordEncoder> encoders = new HashMap<>();
    encoders.put("bcrypt", new BCryptPasswordEncoder());
    encoders.put(encodingId, new SCryptPasswordEncoder());
    return new DelegatingPasswordEncoder(encodingId, encoders);
}

春季版：

  .   ____          _            __ _ _
 /\\ / ___'_ __ _ _(_)_ __  __ _ \ \ \ \
( ( )\___ | '_ | '_| | '_ \/ _` | \ \ \ \
 \\/  ___)| |_)| | | | | || (_| |  ) ) ) )
  '  |____| .__|_| |_|_| |_\__, | / / / /
 =========|_|==============|___/=/_/_/_/
 :: Spring Boot ::       (v1.5.18.RELEASE)

谢谢！

4

1 回答 1

1

您提供的不是有效的 scrypt 哈希

{scrypt}$e0801$QAqC0fvhY....

这是正确的格式（只需替换{scrypt}为$s0）

$s0$e0801$QAqC0fvhY6iJPysiQsFnrcUg205njHo/6o+IDXDn33lxmZOCVBhb4NAqdafhuGmykCxQtMI5xP5zb7MYMUrU3Q==$sBeXCHOm6zQuGdSDKs+HeXnNQGg3bhRidmL+HU/ZTMM=

如果您尝试使用Password4j获取配置：

SCryptFunction.getInstanceFromHash("$s0$e0801$QAqC0...");

你得到N=16384和。在您的 Python 实现中使用此配置。r=8p=1

于 2020-03-24T12:59:11.063 回答