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Java RSA到PHP phpseclib RSA
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Stack Overflow用户
提问于 2016-12-16 23:14:17
回答 1查看 686关注 0票数 0

我正在开发一个支付网关,他们有一个Java演示程序,但是我想用php来实现它。

支付网关使用带有随机生成密钥的3 3DES加密有效负载。该密钥使用支付网关的公钥使用RSA加密。

问题是,当我使用php脚本对该密钥进行RSA加密时,支付网关无法正确提取密钥,显然PHP上的RSA加密不能正常工作…

以下是RSA加密的Java版本:

代码语言:javascript
复制
public static byte[] encrypt(byte[] data, String pubKey64) {

    try {
         byte[] key = Toolkit.base64Decode(pubKey64);
         KeyFactory rsaKeyFac = KeyFactory.getInstance("RSA");
         X509EncodedKeySpec keySpec = new X509EncodedKeySpec(key);
         RSAPublicKey pbk = (RSAPublicKey) rsaKeyFac.generatePublic(keySpec);
         System.out.println("MODE:"+Cipher.ENCRYPT_MODE);
        Cipher cipher = Cipher.getInstance("RSA/ECB/PKCS1PADDING");
        cipher.init(Cipher.ENCRYPT_MODE, pbk);

        byte[] encDate = cipher.doFinal(data);
        return encDate;
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}

在这里,我在PHP脚本中得到了什么:

代码语言:javascript
复制
use phpseclib\Crypt\RSA as RSA;




$PUB_KEY = '-----BEGIN PUBLIC KEY-----
    MIGfMA0GCSqGSIb3DQEBAQUAA4GNADCBiQKBgQDJ1fKGMV/yOUnY1ysFCk0yPP4bfOolC/nTAyHmoser+1yzeLtyYsfitYonFIsXBKoAYwSAhNE+ZSdXZs4A5zt4EKoU+T3IoByCoKgvpCuOx8rgIAqC3O/95pGb9n6rKHR2sz5EPT0aBUUDAB2FJYjA9Sy+kURxa52EOtRKolSmEwIDAQAB
-----END PUBLIC KEY-----';

$PAYLOAD = 'b78850d2f35108b4bc4e7a41';

function encrypt($key,$payload){
    $rsa = new RSA();
    $rsa->loadKey($key); // public key

    $rsa->setEncryptionMode(2);
    $ciphertext = $rsa->encrypt($payload);

    return base64_encode($ciphertext);
}

Java版本使用的是PKCSPADDING,所以我将phpseclib的模式设置为2,即PKCSPADDING,但它仍然不能工作。我有遗漏什么吗?谁能帮我指出一下吗?

更新:

不确定这是否是导致这一现象的原因,但我去掉了“

谢谢大家的帮助。

EN

回答 1

Stack Overflow用户

发布于 2016-12-17 16:20:30

在开始加密过程之前尝试执行define('CRYPT_RSA_PKCS15_COMPAT', true);

引用phpseclib 2.0的RSA.php:

代码语言:javascript
复制
/**
 * RSAES-PKCS1-V1_5-DECRYPT
 *
 * See {@link http://tools.ietf.org/html/rfc3447#section-7.2.2 RFC3447#section-7.2.2}.
 *
 * For compatibility purposes, this function departs slightly from the description given in RFC3447.
 * The reason being that RFC2313#section-8.1 (PKCS#1 v1.5) states that ciphertext's encrypted by the
 * private key should have the second byte set to either 0 or 1 and that ciphertext's encrypted by the
 * public key should have the second byte set to 2.  In RFC3447 (PKCS#1 v2.1), the second byte is supposed
 * to be 2 regardless of which key is used.  For compatibility purposes, we'll just check to make sure the
 * second byte is 2 or less.  If it is, we'll accept the decrypted string as valid.
 *
 * As a consequence of this, a private key encrypted ciphertext produced with \phpseclib\Crypt\RSA may not decrypt
 * with a strictly PKCS#1 v1.5 compliant RSA implementation.  Public key encrypted ciphertext's should but
 * not private key encrypted ciphertext's.
 *
 * @access private
 * @param string $c
 * @return string
 */
票数 1
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/41193731

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