Java的RSA加密

本文介绍了Java的RSA加密 - 解密.NET的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我在Java中使用RSA和尝试使用.NET来解密加密。我包括我的JAVA code和.NET code，希望有人具有这样的事情一定的经验。

JAVA code：

字节[] modulusBytes = Base64.de$c$c("xTSiS4+I/x9awUXcF66Ffw7tracsQfGCn6g6k/hGkLquHYMFTCYk4mOB5NwLwqczwvl8HkQfDShGcvrm47XHKUzA8iadWdA5n4toBECzRxiCWCHm1KEg59LUD3fxTG5ogGiNxDj9wSguCIzFdUxBYq5ot2J4iLgGu0qShml5vwk="); byte []的exponentBytes = Base64.de code（AQAB）; BigInteger的模数=新的BigInteger（1，modulusBytes）; BigInteger的指数=新的BigInteger（1，exponentBytes）; RSAPublicKeySpec rsaPubKey =新RSAPublicKeySpec（模数，指数）; 事实上的KeyFactory = KeyFactory.getInstance（RSA）; 公钥PUBKEY = fact.generatePublic（rsaPubKey）; 密密码= Cipher.getInstance（RSA）; cipher.init（Cipher.ENCRYPT_MODE，PUBKEY）; byte []的plainBytes =新的String（大猫咪舞）的GetBytes（UTF-8）。 字节[]的CipherData = cipher.doFinal（plainBytes）; 字符串encryptedString = Base64.en codeBytes（的CipherData）;

从这个JAVA code我拿encryptedString的恰好是结果：

FoP4+AAIH6hcabXnrvNG5YUk/nBv9n9HU0CAgZjkIWQIDjbOpSwoPVBFERrZ6641x2QaoJw5yv18XAay+0WrCaSw4sveRX+hmPm5qeVUPcjoR4slsVZ/hBFJtAHj9tva4hOugWDZa9s3RVJlxkNfE+u+Kt/YKLOi2EYbH05HjeM=

和尝试使用以下.NET code解密

const int的PROVIDER_RSA_FULL = 1; 常量字符串CONTAINER_NAME =追踪器; CspParameters cspParams; cspParams =新CspParameters（PROVIDER_RSA_FULL）; cspParams.KeyContainerName = CONTAINER_NAME; 的RSACryptoServiceProvider RS​​A1 =新的RSACryptoServiceProvider（cspParams）; rsa1.FromXmlString("<RSAKeyValue><Modulus>xTSiS4+I/x9awUXcF66Ffw7tracsQfGCn6g6k/hGkLquHYMFTCYk4mOB5NwLwqczwvl8HkQfDShGcvrm47XHKUzA8iadWdA5n4toBECzRxiCWCHm1KEg59LUD3fxTG5ogGiNxDj9wSguCIzFdUxBYq5ot2J4iLgGu0qShml5vwk=</Modulus><Exponent>AQAB</Exponent><P>+lXMCEwIN/7+eMpBrq87kQppxu3jJBTwztGTfXNaPUTx+A6uqRwug5oHBbSpYXKNDNCBzVm/0VxB3bo4FJx+ZQ==</P><Q>yasOGaJaE9xlF9T2xRuKeG9ZxCiyjhYaYB/mbtL+SIbtkRLi/AxaU4g2Il/UxhxhSXArKxIzV28zktispPJx1Q==</Q><DP>ueRgQIEFUV+fY979a1RgrVHIPpqEI1URhOMH3Q59oiXCcOumM5njyIHmWQxRAzXnG+7xlKXi1PrnRll0L4oOKQ==</DP><DQ>dfEMNgG1HJhwpxdtmqkYuoakwQvsIRzcIAuIAJh1DoWaupWJGk8/JEstHb1d+t7uJrzrAi2KyT/HscH2diE0YQ==</DQ><InverseQ>YoYF9PF6FiC0YngVeaC/eqt/ea8wMYNN3YO1LuzWpcy2exPRj2U0ZbWMvHXMUb4ea2qmhZGx1QlK4ULAuWKpXQ==</InverseQ><D>g1WAWI4pEK9TA7CA2Yyy/2FzzNiu0uQCuE2TZYRNiomo96KQXpxwqAzZLw+VDXfJMypwDMAVZe/SqzSJnFEtZxjdxaEo3VLcZ1mnbIL0vS7D6iFeYutF9kF231165qGd3k2tgymNMMpY7oYKjS11Y6JqWDU0WE5hjS2X35iG6mE=</D></RSAKeyValue>"); 串data2Decrypt = "FoP4+AAIH6hcabXnrvNG5YUk/nBv9n9HU0CAgZjkIWQIDjbOpSwoPVBFERrZ6641x2QaoJw5yv18XAay+0WrCaSw4sveRX+hmPm5qeVUPcjoR4slsVZ/hBFJtAHj9tva4hOugWDZa9s3RVJlxkNfE+u+Kt/YKLOi2EYbH05HjeM="; byte []的encyrptedBytes = Convert.FromBase64String（da​​ta2Decrypt）; byte []的纯= rsa1.Decrypt（encyrptedBytes，假）; 字符串decryptedString = System.Text.Encoding.UTF8.GetString（普通）;

解决方案

在Java方面，你需要使用RSA / ECB / PKCS1Padding的算法名。随着平原RSA的名字，Java不添加的 PKCS＃1 填充其中C＃实现所期望的，因而有坏数据。

的填充是一个变换输入数据（你烯$ C $光盘字符串）的成具有相同的长度比RSA模（128个字节在这里）稍大串。它是安全的重要，并注入一些随机性。（相同的输入字符串每次都不会产生相同的加密字符串，但解密过程中删除了随机性和仍然恢复正确的字符串）

I am encrypting on JAVA using RSA and attempting to decrypt using .NET. I am including my JAVA code and .NET code in hopes that someone has some experience with this sort of thing.

JAVA Code:

byte[] modulusBytes = Base64.decode("xTSiS4+I/x9awUXcF66Ffw7tracsQfGCn6g6k/hGkLquHYMFTCYk4mOB5NwLwqczwvl8HkQfDShGcvrm47XHKUzA8iadWdA5n4toBECzRxiCWCHm1KEg59LUD3fxTG5ogGiNxDj9wSguCIzFdUxBYq5ot2J4iLgGu0qShml5vwk="); byte[] exponentBytes = Base64.decode("AQAB"); BigInteger modulus = new BigInteger(1, modulusBytes ); BigInteger exponent = new BigInteger(1, exponentBytes); RSAPublicKeySpec rsaPubKey = new RSAPublicKeySpec(modulus, exponent); KeyFactory fact = KeyFactory.getInstance("RSA"); PublicKey pubKey = fact.generatePublic(rsaPubKey); Cipher cipher = Cipher.getInstance("RSA"); cipher.init(Cipher.ENCRYPT_MODE, pubKey); byte[] plainBytes = new String("big kitty dancing").getBytes("UTF-8"); byte[] cipherData = cipher.doFinal( plainBytes ); String encryptedString = Base64.encodeBytes(cipherData);

From this JAVA Code I take the results of the encryptedString which happen to be:

FoP4+AAIH6hcabXnrvNG5YUk/nBv9n9HU0CAgZjkIWQIDjbOpSwoPVBFERrZ6641x2QaoJw5yv18XAay+0WrCaSw4sveRX+hmPm5qeVUPcjoR4slsVZ/hBFJtAHj9tva4hOugWDZa9s3RVJlxkNfE+u+Kt/YKLOi2EYbH05HjeM=

And attempt to decrypt using the following .NET code

const int PROVIDER_RSA_FULL = 1; const string CONTAINER_NAME = "Tracker"; CspParameters cspParams; cspParams = new CspParameters(PROVIDER_RSA_FULL); cspParams.KeyContainerName = CONTAINER_NAME; RSACryptoServiceProvider rsa1 = new RSACryptoServiceProvider(cspParams); rsa1.FromXmlString("<RSAKeyValue><Modulus>xTSiS4+I/x9awUXcF66Ffw7tracsQfGCn6g6k/hGkLquHYMFTCYk4mOB5NwLwqczwvl8HkQfDShGcvrm47XHKUzA8iadWdA5n4toBECzRxiCWCHm1KEg59LUD3fxTG5ogGiNxDj9wSguCIzFdUxBYq5ot2J4iLgGu0qShml5vwk=</Modulus><Exponent>AQAB</Exponent><P>+lXMCEwIN/7+eMpBrq87kQppxu3jJBTwztGTfXNaPUTx+A6uqRwug5oHBbSpYXKNDNCBzVm/0VxB3bo4FJx+ZQ==</P><Q>yasOGaJaE9xlF9T2xRuKeG9ZxCiyjhYaYB/mbtL+SIbtkRLi/AxaU4g2Il/UxhxhSXArKxIzV28zktispPJx1Q==</Q><DP>ueRgQIEFUV+fY979a1RgrVHIPpqEI1URhOMH3Q59oiXCcOumM5njyIHmWQxRAzXnG+7xlKXi1PrnRll0L4oOKQ==</DP><DQ>dfEMNgG1HJhwpxdtmqkYuoakwQvsIRzcIAuIAJh1DoWaupWJGk8/JEstHb1d+t7uJrzrAi2KyT/HscH2diE0YQ==</DQ><InverseQ>YoYF9PF6FiC0YngVeaC/eqt/ea8wMYNN3YO1LuzWpcy2exPRj2U0ZbWMvHXMUb4ea2qmhZGx1QlK4ULAuWKpXQ==</InverseQ><D>g1WAWI4pEK9TA7CA2Yyy/2FzzNiu0uQCuE2TZYRNiomo96KQXpxwqAzZLw+VDXfJMypwDMAVZe/SqzSJnFEtZxjdxaEo3VLcZ1mnbIL0vS7D6iFeYutF9kF231165qGd3k2tgymNMMpY7oYKjS11Y6JqWDU0WE5hjS2X35iG6mE=</D></RSAKeyValue>"); string data2Decrypt = "FoP4+AAIH6hcabXnrvNG5YUk/nBv9n9HU0CAgZjkIWQIDjbOpSwoPVBFERrZ6641x2QaoJw5yv18XAay+0WrCaSw4sveRX+hmPm5qeVUPcjoR4slsVZ/hBFJtAHj9tva4hOugWDZa9s3RVJlxkNfE+u+Kt/YKLOi2EYbH05HjeM="; byte[] encyrptedBytes = Convert.FromBase64String(data2Decrypt); byte[] plain = rsa1.Decrypt(encyrptedBytes, false); string decryptedString = System.Text.Encoding.UTF8.GetString(plain);

解决方案

On the Java side, you need to use "RSA/ECB/PKCS1Padding" as algorithm name. With the plain "RSA" name, Java does not add the PKCS#1 padding which the C# implementation expects, hence the "bad data".

The padding is a transformation of the input data (your encoded string) into a somewhat larger string which has the same length than the RSA modulus (128 bytes here). It is important for security and injects some randomness (the same input string will not yield the same encrypted string every time; but the decryption process removes that randomness and recovers the right string nonetheless).