Golang中的AES加密和Java中的解密

Question

我有以下用Golang編寫的AES加密函數。

func encrypt(key []byte, text string) string {
    plaintext := []byte(text)

    block, err := aes.NewCipher(key)
    if err != nil {
        panic(err)
    }

    ciphertext := make([]byte, aes.BlockSize+len(plaintext))
    iv := ciphertext[:aes.BlockSize]
    if _, err := io.ReadFull(rand.Reader, iv); err != nil {
        panic(err)
    }

    stream := cipher.NewCFBEncrypter(block, iv)
    stream.XORKeyStream(ciphertext[aes.BlockSize:], plaintext)

    return base64.URLEncoding.EncodeToString(ciphertext)
}

我正在努力理解使用Java解密生成的文本的流程。 任何幫助將非常感謝！

這是Scala代碼，不確定它的問題是什么。

def decode(input:String) = {
    val keyBytes = Hex.decodeHex("someKey".toCharArray)
    val inputWithoutPadding = input.substring(0,input.size - 2)
    val inputArr:Seq[Byte] = Hex.decodeHex(inputWithoutPadding.toCharArray)

    val skSpec = new SecretKeySpec(keyBytes, "AES")
    val iv = new IvParameterSpec(inputArr.slice(0,16).toArray)
    val dataToDecrypt = inputArr.slice(16,inputArr.size)

    val cipher = Cipher.getInstance("AES/CFB/NoPadding")
    cipher.init(Cipher.DECRYPT_MODE, skSpec, iv)
    cipher.doFinal(dataToDecrypt.toArray)
}

Answer 1

Java解碼器（另見在線runnable演示 ，打開並單擊“執行”）：

String decode(String base64Text, byte[] key) throws NoSuchPaddingException, NoSuchAlgorithmException, InvalidAlgorithmParameterException, InvalidKeyException, BadPaddingException, IllegalBlockSizeException {
    byte[] inputArr = Base64.getUrlDecoder().decode(base64Text);
    SecretKeySpec skSpec = new SecretKeySpec(key, "AES");
    Cipher cipher = Cipher.getInstance("AES/CFB/NoPadding");
    int blockSize = cipher.getBlockSize();
    IvParameterSpec iv = new IvParameterSpec(Arrays.copyOf(inputArr, blockSize));
    byte[] dataToDecrypt = Arrays.copyOfRange(inputArr, blockSize, inputArr.length);
    cipher.init(Cipher.DECRYPT_MODE, skSpec, iv);
    byte[] result = cipher.doFinal(dataToDecrypt);
    return new String(result, StandardCharsets.UTF_8);
}

Kevin在評論中提供了原始Go編碼器的演示 ，我們可以看到以下結果：

encrypt([]byte("0123456789abcdef"), "test text 123")

是c1bpFhxn74yzHQs-vgLcW6E5yL8zJfgceEQgYl0= 。

讓我們看看上面的Java解碼器如何處理該輸入：

String text = "c1bpFhxn74yzHQs-vgLcW6E5yL8zJfgceEQgYl0=";
byte[] key = "0123456789abcdef".getBytes();
System.out.println(decode(text, key));

打印test text 123 123✔

---

Scala版本（ 在線runnable演示 ）：

def decode(input:String, key:String) = {
    val cipher = Cipher.getInstance("AES/CFB/NoPadding")
    val blockSize = cipher.getBlockSize()
    val keyBytes = key.getBytes()
    val inputArr = Base64.getUrlDecoder().decode(input)
    val skSpec = new SecretKeySpec(keyBytes, "AES")
    val iv = new IvParameterSpec(inputArr.slice(0, blockSize).toArray)
    val dataToDecrypt = inputArr.slice(blockSize, inputArr.size)
    cipher.init(Cipher.DECRYPT_MODE, skSpec, iv)
    new String(cipher.doFinal(dataToDecrypt.toArray))
}

def main(args: Array[String]) {
    print(decode("c1bpFhxn74yzHQs-vgLcW6E5yL8zJfgceEQgYl0=", "0123456789abcdef"));
}

我認為Scala版本中唯一的錯誤是使用Hex.decodeHex 。 您需要一個Base64解碼器，它使用RFC 4648中描述的URL安全字母表， java.util.Base64提供了（自Java 8以來）其getUrlDecoder() 。