解密 Java AWS SDK 中的数据密钥产生乱码
Decrypting data key in Java AWS SDK yields gibberish
所以我一直在阅读和阅读示例,然后......惨遭失败。这是我的情况:
我在 KMS 中有一个 CMK,我已经生成了一个数据密钥,如下所示:
$ aws kms generate-data-key --key-id 64a62e3e-7e38-4f86-8ef2-3d00929e6260 --key-spec AES_256
{
"Plaintext": "+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA=",
"KeyId": "arn:aws:kms:us-west-2:040512153658:key/64a62e3e-7e38-4f86-8ef2-3d00929e6260",
"CiphertextBlob": "AQIDAHgrvfqfgn9D0tTUJOISzFCz7ejMPZ6/HGX0kGAlzKYZ7wEiyHdpuGaOjpq4UQazPAgeAAAAfjB8BgkqhkiG9w0BBwagbzBtAgEAMGgGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMU5JtbI6lxLOv/p4KAgEQgDsX97Pk+ywqLU2VymLRgDSz0exOyzRgLMgd7WEf3sLUh4GnbYllIrxNSdK/DSZrYUhBo78KYugnkTj89g=="
}
然后我通过从 CLI 解密来验证它:
$ aws kms decrypt --ciphertext-blob fileb://<(echo 'AQIDAHgrvfqfgn9D0tTUJOISzFCz7ejMPZ6/HGX0kGAlzKYZ7wEiyHdpuGaOjpq4UQazPAgeAAAAfjB8BgkqhkiG9w0BBwagbzBtAgEAMGgGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMU5JtbI6lxLOv/p4KAgEQgDsX97Pk+ywqLU2VymLRgDSz0exOyzRgLMgd7WEf3sLUh4GnbYllIrxNSdK/DSZrYUhBo78KYugnkTj89g==' | base64 -d) --query Plaintext
"+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA="
瞧瞧!我得到了 Plaintext 值,一切都很好,很干净。然后,我尝试使用 Java 和以下代码通过 SDK 研磨相同的密文 blob:
.
.
.
final String encryptedCipherText = "AQIDAHgrvfqfgn9D0tTUJOISzFCz7ejMPZ6/HGX0kGAlzKYZ7wEiyHdpuGaOjpq4UQazPAgeAAAAfjB8BgkqhkiG9w0BBwagbzBtAgEAMGgGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMU5JtbI6lxLOv/p4KAgEQgDsX97Pk+ywqLU2VymLRgDSz0exOyzRgLMgd7WEf3sLUh4GnbYllIrxNSdK/DSZrYUhBo78KYugnkTj89g==";
final String expectedPlainText = "+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA=";
AWSKMS kmsClient;
String returnValue;
kmsClient = AWSKMSClientBuilder
.standard()
.withRegion("us-west-2")
.build();
ByteBuffer cipherTextBlob = ByteBuffer.wrap(Base64.getDecoder().decode(encryptedCipherText));
DecryptRequest decryptRequest = new DecryptRequest().withCiphertextBlob(cipherTextBlob);
ByteBuffer key = kmsClient.decrypt(decryptRequest).getPlaintext();
final byte[] bytes = new byte[key.remaining()];
key.duplicate().get(bytes);
String result = new String(bytes);
if (expectedPlainText.equals(result)) {
LOG.info("decrypted plaintext matches expected");
} else {
LOG.error("decrypted plaintext unexpected value: " + result);
}
.
.
.
转出的 LOG 条目是:
23:08:33.210 [main] ERROR com.eyefinity.magicmissile.aws.AwsClientConfig - decrypted plaintext unexpected value: �(�k�2!9�5��64=���3o6q�-��
我尝试使用我可用的每个字符集对结果进行编码,但没有字符集生成我的原始明文密钥。就我所见的所有示例而言,我的代码是正确的。那么我做错了什么或者我在这里错过了什么?我想要的只是以一个 Java 字符串变量结束,其中包含 "+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA=".
我偶然发现了自己的解决方案:我太接近了!在上面的代码中提取从 KMS 返回的相同明文值以及我在生成数据密钥时在 aws-cli 命令行上收到的 ASCII 字符串所需的全部是获取字节数组并对其进行 Base64 编码。所以参考我上面的示例代码,一直在顶部,替换读取的行...
String result = new String(bytes);
像这样:
String result = Base64.getEncoder().encodeToString(bytes);
所以我一直在阅读和阅读示例,然后......惨遭失败。这是我的情况:
我在 KMS 中有一个 CMK,我已经生成了一个数据密钥,如下所示:
$ aws kms generate-data-key --key-id 64a62e3e-7e38-4f86-8ef2-3d00929e6260 --key-spec AES_256
{
"Plaintext": "+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA=",
"KeyId": "arn:aws:kms:us-west-2:040512153658:key/64a62e3e-7e38-4f86-8ef2-3d00929e6260",
"CiphertextBlob": "AQIDAHgrvfqfgn9D0tTUJOISzFCz7ejMPZ6/HGX0kGAlzKYZ7wEiyHdpuGaOjpq4UQazPAgeAAAAfjB8BgkqhkiG9w0BBwagbzBtAgEAMGgGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMU5JtbI6lxLOv/p4KAgEQgDsX97Pk+ywqLU2VymLRgDSz0exOyzRgLMgd7WEf3sLUh4GnbYllIrxNSdK/DSZrYUhBo78KYugnkTj89g=="
}
然后我通过从 CLI 解密来验证它:
$ aws kms decrypt --ciphertext-blob fileb://<(echo 'AQIDAHgrvfqfgn9D0tTUJOISzFCz7ejMPZ6/HGX0kGAlzKYZ7wEiyHdpuGaOjpq4UQazPAgeAAAAfjB8BgkqhkiG9w0BBwagbzBtAgEAMGgGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMU5JtbI6lxLOv/p4KAgEQgDsX97Pk+ywqLU2VymLRgDSz0exOyzRgLMgd7WEf3sLUh4GnbYllIrxNSdK/DSZrYUhBo78KYugnkTj89g==' | base64 -d) --query Plaintext
"+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA="
瞧瞧!我得到了 Plaintext 值,一切都很好,很干净。然后,我尝试使用 Java 和以下代码通过 SDK 研磨相同的密文 blob:
.
.
.
final String encryptedCipherText = "AQIDAHgrvfqfgn9D0tTUJOISzFCz7ejMPZ6/HGX0kGAlzKYZ7wEiyHdpuGaOjpq4UQazPAgeAAAAfjB8BgkqhkiG9w0BBwagbzBtAgEAMGgGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMU5JtbI6lxLOv/p4KAgEQgDsX97Pk+ywqLU2VymLRgDSz0exOyzRgLMgd7WEf3sLUh4GnbYllIrxNSdK/DSZrYUhBo78KYugnkTj89g==";
final String expectedPlainText = "+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA=";
AWSKMS kmsClient;
String returnValue;
kmsClient = AWSKMSClientBuilder
.standard()
.withRegion("us-west-2")
.build();
ByteBuffer cipherTextBlob = ByteBuffer.wrap(Base64.getDecoder().decode(encryptedCipherText));
DecryptRequest decryptRequest = new DecryptRequest().withCiphertextBlob(cipherTextBlob);
ByteBuffer key = kmsClient.decrypt(decryptRequest).getPlaintext();
final byte[] bytes = new byte[key.remaining()];
key.duplicate().get(bytes);
String result = new String(bytes);
if (expectedPlainText.equals(result)) {
LOG.info("decrypted plaintext matches expected");
} else {
LOG.error("decrypted plaintext unexpected value: " + result);
}
.
.
.
转出的 LOG 条目是:
23:08:33.210 [main] ERROR com.eyefinity.magicmissile.aws.AwsClientConfig - decrypted plaintext unexpected value: �(�k�2!9�5��64=���3o6q�-��
我尝试使用我可用的每个字符集对结果进行编码,但没有字符集生成我的原始明文密钥。就我所见的所有示例而言,我的代码是正确的。那么我做错了什么或者我在这里错过了什么?我想要的只是以一个 Java 字符串变量结束,其中包含 "+SjeaxtD5TIhOcY16+A2NA493MbxnYozbzZx4i3/BfA=".
我偶然发现了自己的解决方案:我太接近了!在上面的代码中提取从 KMS 返回的相同明文值以及我在生成数据密钥时在 aws-cli 命令行上收到的 ASCII 字符串所需的全部是获取字节数组并对其进行 Base64 编码。所以参考我上面的示例代码,一直在顶部,替换读取的行...
String result = new String(bytes);
像这样:
String result = Base64.getEncoder().encodeToString(bytes);