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如何在Java中从PGP公钥中获取用户id?

pgp gnupg apache-camel java

Gangaraju Anjali Pavithra · 技术社区 · 7 年前

PGPDataFormat pgpDataFormat=new PGPDataFormat();
pgpDataFormat.setKeyFileName("0x6E1A09A4-pub.asc");
pgpDataFormat.setKeyUserid("user@domain.com");
pgpDataFormat.marshal(exchange, exchange.getIn().getBody(File.class), exchange.getIn().getBody(OutputStream.class));

KeyUserId 和公钥。我想从公钥中提取这个用户id。

$ gpg --import 0x6E1A09A4-pub.asc                    

gpg: key 6E1A09A4: public key "User <user@domain.com>" imported
gpg: Total number processed: 1
gpg:               imported: 1  (RSA: 1)

如果我使用 gpg

1 回复 | 直到 7 年前

Siva 7 年前

private PGPPublicKey getPGPPublicKey(String filePath) throws IOException, PGPException {
    InputStream inputStream = new BufferedInputStream(new FileInputStream(filePath));
    PGPPublicKeyRingCollection pgpPublicKeyRingCollection = new PGPPublicKeyRingCollection(PGPUtil.getDecoderStream(inputStream), new JcaKeyFingerprintCalculator());
    Iterator keyRingIterator = pgpPublicKeyRingCollection.getKeyRings();
    while (keyRingIterator.hasNext()) {
        PGPPublicKeyRing keyRing = (PGPPublicKeyRing) keyRingIterator.next();
        Iterator keyIterator = keyRing.getPublicKeys();
        while (keyIterator.hasNext()) {
            PGPPublicKey key = (PGPPublicKey) keyIterator.next();
            if (key.isEncryptionKey()) {
                return key;
            }
        }
    }
    inputStream.close();
    throw new IllegalArgumentException("Can't find encryption key in key ring.");
}

private String getUserId(String publicKeyFile) throws IOException, PGPException {
    PGPPublicKey pgpPublicKey = getPGPPublicKey(publicKeyFile);
    // You can iterate and return all the user ids if needed
    return (String) pgpPublicKey.getUserIDs().next();
}