Question

我正在验证来自Azure的JWT令牌并使用JJWT。我从与我的tid相关的密钥文档中检索模数和指数，这些是字段 n和e分别。验证失败并出现错误：JWT签名与本地计算的签名不匹配。 JWT有效性不能被断言，也不应该被信任。

这是代码。有人看到我犯的错误吗？代码运行正常，直到它抛出签名不匹配错误的验证。

private Claims extractClaimsForRsaSignedJwts(String token, String mod, String exp) {
    Claims claims = null;
    byte[] modBytes = Base64.decodeBase64(mod.getBytes());
    byte[] expBytes = Base64.decodeBase64(exp.getBytes());
    BigInteger modulus = new BigInteger(modBytes);
    BigInteger exponent = new BigInteger(expBytes);
    RSAPublicKeySpec pubKeySpecification = new RSAPublicKeySpec(modulus, exponent);
    KeyFactory keyFac = null;
    try {
        keyFac = KeyFactory.getInstance("RSA");
    } catch (NoSuchAlgorithmException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }
    RSAPublicKey rsaPub = null;
    try {
        rsaPub = (RSAPublicKey) keyFac.generatePublic(pubKeySpecification);
    } catch (Exception e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }

    JwtParser jwtParser = Jwts.parser().setSigningKey(rsaPub);

    try {
        claims = jwtParser.parseClaimsJws(token).getBody();
    } catch (Exception e) {
        // JWT signature does not match locally computed signature. JWT validity cannot be asserted and should not be trusted.
        System.out.println("The RSA JWT key validation failed: " + e.getMessage());
    }

    return claims;
}

谢谢！

扬

Answer 1

我发现了问题！对于正数，BigInteger应该用signum 1构造！现在，代码就像AzureAD JWT签名验证的魅力一样。

    BigInteger modulus  = new BigInteger(1, modBytes);
    BigInteger exponent = new BigInteger(1, expBytes);

这是最终的代码： Screen Shot Of Code With Correction

如何生成RSAPublicKey以提供给JJWT RSA令牌验证

1 个答案: