Question

我有这个方法：

public static PublicKey getKey(String key){
    try{
        byte[] byteKey = Base64.getDecoder().decode(key.getBytes());
        X509EncodedKeySpec X509publicKey = new X509EncodedKeySpec(byteKey);
        KeyFactory kf = KeyFactory.getInstance("RSA");

        return kf.generatePublic(X509publicKey);
    }
    catch(Exception e){
        e.printStackTrace();
    }

        return null;
    }
}

这应该转换一个公钥（此JSON中的n个属性之一来自Salesforce public key endpoint）。在我的钥匙应该是孩子208，但是：

getKey("pTSWJvy04hAU7ev7wmaTvpwHsEbseuPl0AlwoxPHnmoOIMATRT0eTqYpLJxDp4BHRFxDTrcUKHKVHGIAVut_-l6nkEI6ALOVW9C5PP9bXwqeHJ5tiGA6AMpaY1LsJJOd2lgExr0LHUPF2TtO4LOVmlptyGPTRSWhmRpPSc5bjGFsyTFr78WmeixjEts9icAUCiBDdpwVw9qVdjJPsufyimqp8os5htm-DB_qKsnRwABVSQRKLw2y7Mr7NP31R07Mpr108dLS5Et8tKnFoiX0MHf5BbA50NG3DtyT27nSa14T0IvSWYJQhxPWXYtuXkVldpbZZn8E2B1VpcDXXGnSPXoPstE9BU7w1RaF31UeEsE8sWDFdHgJMUYMfWjxVhdJB_NaVUt7v0X9QPffXcFPWKUUeVx0g_ONMz0bB_HltFSODWJeAvcZXk14u4sjpKXNIiK8dJzMe0Qn10m5nIdVtxzXfKHiDNobX1dMENDplhDMVk1eYS9x8dh90qt68Q59vKusm6iMjSaXtzwMfqJBwNck41e9Pie2m_cdSu-RIq1u3FScGkezzqNvIuzcse2y-6ApHkkaFMVsNw3CbD87LtVgNusuom38UVC1rS0LeLYFinU3hnFwvB1UCv1_0Cs8CmrOmdifqd25aZIDf5p8f3kowq3QY1mNtdQc9-HgZ3k");

给我一个错误：

java.lang.IllegalArgumentException：非法的base64字符5f at java.util.Base64 $ Decoder.decode0（Base64.java:714）at java.util.Base64 $ Decoder.decode（Base64.java:526）at testJwt.testJwt.App.getKey（App.java:52）at testJwt.testJwt.App.main（App.java:76）

我该怎么办？我做错了什么？

Answer 1

您拥有的数据不是完整的RSA公钥，而只是模数，或者＆＃34; n＆＃34;值。＆＃34; e＆＃34; value是公钥的另一部分。这是代码片段来说明它。

import java.math.BigInteger;
import java.security.KeyFactory;
import java.security.PublicKey;
import java.security.spec.RSAPublicKeySpec;
import java.util.Base64;

public class Main {

    public static void main(String[] args) throws Exception {
        String jsonN = "pTSWJvy04hAU7ev7wmaTvpwHsEbseuPl0AlwoxPHnmoOIMATRT0eTqYpLJxDp4BHRFxDTrcUKHKVHGIA" +
                "Vut_-l6nkEI6ALOVW9C5PP9bXwqeHJ5tiGA6AMpaY1LsJJOd2lgExr0LHUPF2TtO4LOVmlptyGPTRSWhmRpPSc5" +
                "bjGFsyTFr78WmeixjEts9icAUCiBDdpwVw9qVdjJPsufyimqp8os5htm-DB_qKsnRwABVSQRKLw2y7Mr7NP31R0" +
                "7Mpr108dLS5Et8tKnFoiX0MHf5BbA50NG3DtyT27nSa14T0IvSWYJQhxPWXYtuXkVldpbZZn8E2B1VpcDXXGnSP" +
                "XoPstE9BU7w1RaF31UeEsE8sWDFdHgJMUYMfWjxVhdJB_NaVUt7v0X9QPffXcFPWKUUeVx0g_ONMz0bB_HltFSO" +
                "DWJeAvcZXk14u4sjpKXNIiK8dJzMe0Qn10m5nIdVtxzXfKHiDNobX1dMENDplhDMVk1eYS9x8dh90qt68Q59vKu" +
                "sm6iMjSaXtzwMfqJBwNck41e9Pie2m_cdSu-RIq1u3FScGkezzqNvIuzcse2y-6ApHkkaFMVsNw3CbD87LtVgNu" +
                "suom38UVC1rS0LeLYFinU3hnFwvB1UCv1_0Cs8CmrOmdifqd25aZIDf5p8f3kowq3QY1mNtdQc9-HgZ3k";
        String jsonE = "AQAB";

        byte [] nBytes = Base64.getUrlDecoder().decode(jsonN);
        byte [] eBytes = Base64.getUrlDecoder().decode(jsonE);

        BigInteger n = new BigInteger(1, nBytes); // need to use the sign-magnitude constructor
        BigInteger e = new BigInteger(1, eBytes);

        RSAPublicKeySpec rsaPublicKeySpec = new RSAPublicKeySpec(n, e);
        KeyFactory rsaKeyFactory = KeyFactory.getInstance("RSA");
        PublicKey rsaPub = rsaKeyFactory.generatePublic(rsaPublicKeySpec);
        System.out.println(rsaPub);
    }
}

我不会手动执行此操作，而是寻找已经执行此操作的库，尽管库建议在stackoverflow上是偏离主题的。

Answer 2

您似乎在Base64字符串中有_和-（而不是/和+）。然后是URL安全编码的Base64：

Base64.getUrlDecoder().decode(key.getBytes())

从String到java.security.PublicKey对象给我一个非法字符

2 个答案: