他们给我发了一个公钥文件,我用以下命令检查了密钥(结果如下)：

C:\Program Files (x86)\GNU\GnuPG>gpg2 --dry-run -vvvv --import \test\thekey.txt

gpg: using character set `CP437'
gpg: armor: BEGIN PGP PUBLIC KEY BLOCK
gpg: armor header: Version: GnuPG v1.2.6 (GNU/Linux)
:public key packet:
        version 4,algo 17,created 1232463981,expires 0
        pkey[0]: [1024 bits]
        pkey[1]: [160 bits]
        pkey[2]: [1024 bits]
        pkey[3]: [1021 bits]
        keyid: C0B18494F5B25CB6
:user ID packet: "FTP Upload Key 3 <XXXX.XXXX@XXXXXXX.com>"
:signature packet: algo 17,keyid C0B18494F5B25CB6
        version 4,md5len 0,sigclass 0x13
        digest algo 2,begin of digest 23 e8
        hashed subpkt 2 len 4 (sig created 2009-01-20)
        hashed subpkt 27 len 1 (key flags: 03)
        hashed subpkt 11 len 5 (pref-sym-algos: 9 8 7 3 2)
        hashed subpkt 21 len 2 (pref-hash-algos: 2 3)
        hashed subpkt 22 len 2 (pref-zip-algos: 2 1)
        hashed subpkt 30 len 1 (features: 01)
        hashed subpkt 23 len 1 (key server preferences: 80)
        subpkt 16 len 8 (issuer key ID C0B18494F5B25CB6)
        data: [159 bits]
        data: [156 bits]
:public sub key packet:
        version 4,algo 16,expires 0
        pkey[0]: [1024 bits]
        pkey[1]: [4 bits]
        pkey[2]: [1024 bits]
        keyid: 358DC68F4A83D261
:signature packet: algo 17,sigclass 0x18
        digest algo 2,begin of digest 28 08
        hashed subpkt 2 len 4 (sig created 2009-01-20)
        hashed subpkt 27 len 1 (key flags: 0C)
        subpkt 16 len 8 (issuer key ID C0B18494F5B25CB6)
        data: [158 bits]
        data: [160 bits]

gpg: pub  1024D/F5B25CB6 2009-01-20  FTP Upload Key 3 <XXXX.XXXX@XXXXXXX.com>
gpg: writing to `C:/Users/jholovacs/AppData/Roaming/gnupg/pubring.gpg'
gpg: using PGP trust model
gpg: key 9166F91E: accepted as trusted key
gpg: key DBA95ED9: accepted as trusted key
gpg: key DF548FCC: accepted as trusted key
gpg: key F5B25CB6: public key "[User ID not found]" imported
gpg: Total number processed: 1
gpg:               imported: 1

我拿了一个示例文本文件,并使用他们的公钥对其进行加密,然后用这个命令检查了结果文件(结果如下)：

C:\Program Files (x86)\GNU\GnuPG>gpg2 --list-packets <\test\bankfile.pgp
:compressed packet: algo=1
:pubkey enc packet: version 3,keyid 358DC68F4A83D261
        data: [1019 bits]
        data: [1024 bits]
:encrypted data packet:
        length: 620
gpg: encrypted with ELG key,ID 4A83D261
gpg: decryption Failed: No secret key

事情是这样的：当我将文件上传到我们的银行时,他们说文件与他们的私钥不匹配.我看到358DC68F4A83D261的密钥id与他们的公钥相匹配,所以我摸不着头……但我也看到ID为4A83D261的ElGamal密钥似乎没有反映在他们的公钥文件中.

我正在使用基于BouncyCastle库的自定义加密解决方案,因此加密时可能出现错误(我不能使用我自己的Pgp4Win版本复制错误).它们也可能存在问题.我的问题是,我不确定如何阅读这些反馈以确定问题所在.一些建议将不胜感激.