java.security.InvalidKeyException : Parameters missing

Question

我需要将加密密码存储在数据库中而不存储任何 key 或盐。我想确保它也更安全，即使它是双向加密。因此，在谷歌搜索了一下之后，我创建了一个示例程序来测试它。我的想法是创建一个自定义 JPA AttributeConverter 类来管理它。

程序如下:

import java.security.Key;
import java.security.spec.KeySpec;
import java.util.Base64;

import javax.crypto.Cipher;
import javax.crypto.SecretKey;
import javax.crypto.SecretKeyFactory;
import javax.crypto.spec.PBEKeySpec;
import javax.crypto.spec.SecretKeySpec;

public class Crypto {

    private static final String _algorithm = "AES";
    private static final String _password = "_pasword*";
    private static final String _salt = "_salt*";
    private static final String _keygen_spec = "PBKDF2WithHmacSHA1";
    private static final String _cipher_spec = "AES/CBC/PKCS5Padding";

    public static String encrypt(String data) throws Exception {
        Key key = getKey();
        System.out.println(key.toString());
        Cipher cipher = Cipher.getInstance(_cipher_spec);
        cipher.init(Cipher.ENCRYPT_MODE, key);
        byte[] encVal = cipher.doFinal(data.getBytes());
        String encryptedValue = Base64.getEncoder().encodeToString(encVal);
        System.out.println("Encrypted value of "+data+": "+encryptedValue);
        return encryptedValue;
    }

    public static void decrypt(String encryptedData) throws Exception {
        Key key = getKey();
        System.out.println(key.toString());
        Cipher cipher = Cipher.getInstance(_cipher_spec);
        cipher.init(Cipher.DECRYPT_MODE, key);
        byte[] decordedValue = Base64.getDecoder().decode(encryptedData);
        byte[] decValue = cipher.doFinal(decordedValue);
        String decryptedValue = new String(decValue);
        System.out.println("Decrypted value of "+encryptedData+": "+decryptedValue);
    }

    private static Key getKey() throws Exception {
        SecretKeyFactory factory = SecretKeyFactory.getInstance(_keygen_spec);
        KeySpec spec = new PBEKeySpec(_password.toCharArray(), _salt.getBytes(), 65536, 128);
        SecretKey tmp = factory.generateSecret(spec);
        SecretKey secret = new SecretKeySpec(tmp.getEncoded(), _algorithm);
        return secret;
    }

    public static void main(String []str) throws Exception {
        String value = encrypt("India@123");
        decrypt(value);
    }
}

但它抛出以下异常:

javax.crypto.spec.SecretKeySpec@17111
Encrypted value of India@123: iAv1fvjMnJqilg90rGztXA==
javax.crypto.spec.SecretKeySpec@17111
Exception in thread "main" java.security.InvalidKeyException: Parameters missing
    at com.sun.crypto.provider.CipherCore.init(CipherCore.java:469)
    at com.sun.crypto.provider.AESCipher.engineInit(AESCipher.java:313)
    at javax.crypto.Cipher.implInit(Cipher.java:802)
    at javax.crypto.Cipher.chooseProvider(Cipher.java:864)
    at javax.crypto.Cipher.init(Cipher.java:1249)
    at javax.crypto.Cipher.init(Cipher.java:1186)
    at org.lp.test.Crypto.decrypt(Crypto.java:37)
    at org.lp.test.Crypto.main(Crypto.java:54)

我无法弄清楚这一点。

我已经根据 @Luke Park 的回答纠正了异常，我创建了一个 JPA AttributeConverter，如下所示:

    import java.security.Key;
import java.security.spec.KeySpec;
import java.util.Base64;

import javax.crypto.Cipher;
import javax.crypto.SecretKey;
import javax.crypto.SecretKeyFactory;
import javax.crypto.spec.PBEKeySpec;
import javax.crypto.spec.SecretKeySpec;
import javax.persistence.AttributeConverter;
import javax.persistence.Converter;

@Converter(autoApply=true)
public class CryptoJPAConverter implements AttributeConverter<String, String> {

    private static final String _algorithm = "AES";
    private static final String _password = "_pasword*";
    private static final String _salt = "_salt*";
    private static final String _keygen_spec = "PBKDF2WithHmacSHA1";
    private static final String _cipher_spec = "AES/ECB/PKCS5Padding";

    @Override
    public String convertToDatabaseColumn(String clearText) {
        Key key;
        Cipher cipher;
        try {
            key = getKey();
            cipher = Cipher.getInstance(_cipher_spec);
            cipher.init(Cipher.ENCRYPT_MODE, key);
            byte[] encVal = cipher.doFinal(clearText.getBytes());
            String encryptedValue = Base64.getEncoder().encodeToString(encVal);
            return encryptedValue;
        } catch (Exception e) {
            throw new RuntimeException(e);
        }
    }

    @Override
    public String convertToEntityAttribute(String encryptedText) {
        Key key;
        try {
            key = getKey();
            Cipher cipher = Cipher.getInstance(_cipher_spec);
            cipher.init(Cipher.DECRYPT_MODE, key);
            byte[] decordedValue = Base64.getDecoder().decode(encryptedText);
            byte[] decValue = cipher.doFinal(decordedValue);
            String decryptedValue = new String(decValue);
            return decryptedValue;
        } catch (Exception e) {
            throw new RuntimeException(e);
        }
    }

    private static Key getKey() throws Exception {
        SecretKeyFactory factory = SecretKeyFactory.getInstance(_keygen_spec);
        KeySpec spec = new PBEKeySpec(_password.toCharArray(), _salt.getBytes(), 65536, 128);
        SecretKey tmp = factory.generateSecret(spec);
        SecretKey secret = new SecretKeySpec(tmp.getEncoded(), _algorithm);
        return secret;
    }
}

我使用了两种方式加密，因为我需要将密码以明文形式传递给 Java 邮件客户端。

欢迎提出建议和意见

Answer 1

您正在使用 AES/CBC/PKCS5Padding，但您没有将 IV 传递给 cipher.init 调用。

CBC 模式要求每次加密操作都有一个随机 IV，您应该在每次加密时使用 SecureRandom 生成此值，并将该值作为 IvParameterSpec 传入。您将需要相同的 IV 来解密。通常将 IV 添加到密文中并在需要时检索。

另外一点，加密密码确实是一个非常糟糕的主意。您必须问这个问题的事实在某种程度上证明您无法做出与安全相关的决策。为了你自己和你的项目，请散列你的密码。 PBKDF2 和 bcrypt 都是不错的方法。