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如何在Java中解码PKCS#5加密PKCS#8私钥

  •  2
  • L. Cornelius Dol  · 技术社区  · 17 年前

    -----BEGIN ENCRYPTED PRIVATE KEY-----
    MIICmDAaBgkqhkiG9w0BBQMwDQQIybM2XFqx4EwCAQUEggJ4MKg/NE+L6NJgbOf4
    ...
    8QnGu4R7lFlweH/VAK8n0L75h3q2g62MKLJqmKLtAILNve4zymnO+LVZ4Js=
    -----END ENCRYPTED PRIVATE KEY-----
    

    为此,我需要获取一个Java密钥对象,然后将其与匹配的证书一起添加到密钥库中。私钥用100字节的二进制密钥加密。

    证书对象的创建很简单,但我似乎不知道如何从上述Base64编码的PKCS#5密钥转换为解密的PKCS#8 RSA私钥。此时我遇到了障碍,因为SecretKeyFactory.GenerateCret()调用失败,原因是:

    InvalidKeySpecException: Password is not ASCII
    

    谁能告诉我如何从Base64编码的值获取可以添加到密钥库的密钥对象?


    结论

    我随后遇到的问题是,我的PKCS#8数据没有正确编码(这显然是早期SSL实现中的一个常见错误),因为PKCS#1数据需要包装在ASN.1八进制字符串中。我编写了一个简单的补丁例程,它将处理我的密钥,已知密钥长度在512到4096位之间(见下文)。


    私钥解码器

    private PrivateKey readPrivateKey(File inpfil) throws IOException, GeneralSecurityException {
        String[]                            pbeb64s;                                // PBE ASN.1 data base-64 encoded
    
        byte[]                              pbedta;                                 // PBE ASN.1 data in bytes
        EncryptedPrivateKeyInfo             pbeinf;                                 // PBE key info
        PBEParameterSpec                    pbeprm;                                 // PBE parameters
        Cipher                              pbecph;                                 // PBE decryption cipher
    
        byte[]                              pk8dta;                                 // PKCS#8 ASN.1 data in bytes
        KeyFactory                          pk8fac=KeyFactory.getInstance("RSA");   // PKCS#8 key factory for decoding private key from ASN.1 data.
    
        pbeb64s=readDataBlocks(inpfil,"ENCRYPTED PRIVATE KEY");
        if(pbeb64s.length!=1) { throw new GeneralSecurityException("The keystore '"+inpfil+"' contains multiple private keys"); }
        pbedta=base64.decode(pbeb64s[0]);
        log.diagln("  - Read private key data");
    
        pbeinf=new EncryptedPrivateKeyInfo(pbedta);
        pbeprm=(PBEParameterSpec)pbeinf.getAlgParameters().getParameterSpec(PBEParameterSpec.class);
        pbecph=Cipher.getInstance(pbeinf.getAlgName());
        pbecph.init(Cipher.DECRYPT_MODE,pbeDecryptKey,pbeprm);
    
        pk8dta=pbecph.doFinal(pbeinf.getEncryptedData());
        log.diagln("  - Private Key: Algorithm= "+pbeinf.getAlgName()+", Iterations: "+pbeprm.getIterationCount()+", Salt: "+Base16.toString(pbeprm.getSalt()));
        pk8dta=patchKeyData(inpfil,pk8dta);
        return pk8fac.generatePrivate(new PKCS8EncodedKeySpec(pk8dta));
        }
    

    import java.io.*;
    import java.security.*;
    import java.security.spec.*;
    import java.util.*;
    import javax.crypto.*;
    import javax.crypto.spec.*;
    
    class BinaryPBEKey
    extends Object
    implements SecretKey
    {
    private final byte[]                    key;
    
    /**
     * Creates a PBE key from a given binary key.
     *
     * @param key       The key.
     */
    BinaryPBEKey(byte[] key) throws InvalidKeySpecException {
        if(key==null) { this.key=new byte[0];         }
        else          { this.key=(byte[])key.clone(); }
        Arrays.fill(key,(byte)0);
        }
    
    public byte[] getEncoded() {
        return (byte[])key.clone();
        }
    
    public String getAlgorithm() {
        return "PBEWithMD5AndDES";
        }
    
    public String getFormat() {
        return "RAW";
        }
    
    /**
     * Calculates a hash code value for the object.
     * Objects that are equal will also have the same hashcode.
     */
    public int hashCode() {
        int                             ret=0;
    
        for(int xa=1; xa<this.key.length; xa++) { ret+=(this.key[xa]*xa); }
        return (ret^=getAlgorithm().toLowerCase().hashCode());
        }
    
    public boolean equals(Object obj) {
        if(obj==this                 ) { return true;  }
        if(obj.getClass()!=getClass()) { return false; }
    
        BinaryPBEKey                    oth=(BinaryPBEKey)obj;
    
        if(!(oth.getAlgorithm().equalsIgnoreCase(getAlgorithm()))) {
            return false;
            }
    
        byte[]  othkey=oth.getEncoded();
        boolean ret   =Arrays.equals(key,othkey);
        Arrays.fill(othkey,(byte)0);
        return ret;
        }
    
    public void destroy() {
        Arrays.fill(this.key,(byte)0);
        }
    
    /**
     * Ensure that the password bytes of this key are zeroed out when there are no more references to it.
     */
    protected void finalize() throws Throwable {
        try { destroy(); } finally { super.finalize(); }
        }
    

    PKCS#8配线

    /**
     * Patch the private key ASN.1 data to conform to PKCS#8.
     * <p>
     * The SSLPlus private key is not properly encoded PKCS#8 - the PKCS#1 RSAPrivateKey should have been wrapped
     * inside an OctetString, thus:
     * <pre>
     * SSLPlus Encoding:
     *        0 30  627: SEQUENCE {
     *        4 02    1:   INTEGER 0
     *        7 30   13:   SEQUENCE {
     *        9 06    9:     OBJECT IDENTIFIER rsaEncryption (1 2 840 113549 1 1 1)
     *       20 05    0:     NULL
     *                 :     }
     *       22 30  605:   SEQUENCE {
     *       26 02    1:     INTEGER 0
     *       29 02  129:     INTEGER
     *                 :       00 CA 72 B8 D1 B8 8E B9 39 C0 92 C1 4C 53 B4 F4
     *       ...
     *
     * PKCS#8 Encoding
     *       0 30  631: SEQUENCE {
     *       4 02    1:   INTEGER 0
     *       7 30   13:   SEQUENCE {
     *       9 06    9:     OBJECT IDENTIFIER rsaEncryption (1 2 840 113549 1 1 1)
     *      20 05    0:     NULL
     *                :     }
     * ==>  22 04  609:   OCTET STRING, encapsulates {
     *      26 30  605:       SEQUENCE {
     *      30 02    1:         INTEGER 0
     *      33 02  129:         INTEGER
     *                :           00 CA 72 B8 D1 B8 8E B9 39 C0 92 C1 4C 53 B4 F4
     *      ...
     * </pre>
     *
     * Hex Dumps (1K key, space padded for clarity):
     *    Before      : 30 820271 020100300D06092A864886F70D0101010500           30 82025B ... A228
     *    After       : 30 820275 020100300D06092A864886F70D0101010500 04 82025F 30 82025B ... A228
     *                     ^^^^^^                                         ^^^^^^
     *                     Add 4 for later 0482xxxx                       Original total + 4 - 22 (equals the key length of 025B+4)
     */
    private byte[] patchKeyData(File inpfil, byte[] asndta) throws IOException, GeneralSecurityException { // except it really doesn't throw an exception
        ByteArrayOutputStream               patdta=new ByteArrayOutputStream();
        int                                 orglen=decodeAsnLength(inpfil,asndta,1);
    
        patdta.write(asndta,0,1);                                                   // original leader type
        patdta.write(encodeAsnLength(inpfil,(orglen+4)));                           // new total length
        patdta.write(asndta,4,(22-4));                                              // bit between total length an where octet-string wrapper needs to be inserted
        patdta.write(0x04);                                                         // octet-string type
        patdta.write(encodeAsnLength(inpfil,(orglen+4-22)));                        // octet-string length (key data type+key data length+key data)
        patdta.write(asndta,22,asndta.length-22);                                   // private key data
        return patdta.toByteArray();
        }
    
    private int decodeAsnLength(File inpfil, byte[] asndta, int ofs) throws GeneralSecurityException {
        if((asndta[ofs]&0xFF)==0x82) { return (((asndta[ofs+1]&0x000000FF)<< 8)|((asndta[ofs+2]&0x000000FF)));                                                           }
        else                         { throw new GeneralSecurityException("The private key in file '"+inpfil+"' is not supported (ID="+Base16.toString(asndta,0,4)+")"); }
        }
    
    private byte[] encodeAsnLength(File inpfil, int len) throws GeneralSecurityException {
        if(len>=0x0100 && len<=0xFFFF) { return new byte[]{ (byte)0x82,(byte)((len>>>8)&0x000000FF),(byte)len };                                                            }
        else                           { throw new GeneralSecurityException("The new length of "+len+" for patching the private key in file '"+inpfil+"' is out of range"); }
        }
    
    2 回复  |  直到 17 年前
        1
  •  1
  •   Rasmus Faber    17 年前

    我刚刚将您解密的数据转储到ASN.1解析器中,在我看来,ASN.1非常好:

           0 30  627: SEQUENCE {
           4 02    1:   INTEGER 0
           7 30   13:   SEQUENCE {
           9 06    9:     OBJECT IDENTIFIER rsaEncryption (1 2 840 113549 1 1 1)
          20 05    0:     NULL
                    :     }
          22 30  605:   SEQUENCE {
          26 02    1:     INTEGER 0
          29 02  129:     INTEGER
                    :       00 CA 72 B8 D1 B8 8E B9 39 C0 92 C1 4C 53 B4 F4
                    :       38 48 3F C3 1C DC 6B BC BE 26 A3 B2 F7 7C 60 A8
                    :       2C 0D 86 ED FC 2D D2 5C 99 B6 B6 71 A8 6D 2F 51
                    :       25 FA 9C 42 FE 10 C1 2F 39 EA E8 FF 1A 78 BA 6B
                    :       64 B8 39 34 3B F4 1C 45 06 C3 B9 98 DC 01 FF 41
                    :       56 36 4F DD 35 69 A4 27 BB 5F FD DD 5C 73 BA 9A
                    :       94 5A 4F 37 A9 48 3D 5B 89 EA EE BA 8D 02 6E D7
                    :       6E D4 6F BC 7D 7A A4 41 4C 4D CA 08 05 20 66 A3
                    :               [ Another 1 bytes skipped ]
         161 02    3:     INTEGER 65537
         166 02  128:     INTEGER
                    :       21 6A E2 7B 2B DD D3 51 67 2A 52 62 09 07 3B B0
                    :       F6 AC 1F C6 E9 D3 96 EA 44 72 8D 1E 31 17 BB 6A
                    :       DA 28 C5 AB F4 DC 5E 90 B9 0A 50 A4 9E B1 4A D1
                    :       DC 16 63 30 91 0F 72 7E 3A FA 8E F1 8D B0 27 FD
                    :       C2 BA B5 F8 FC 7C 46 C0 FD AD A7 39 7C 36 71 7A
                    :       33 8B AD 0D 0C DA 50 B7 0E BF D8 64 7D 44 BD 64
                    :       6F E2 51 B7 5E 2D 7B BA 02 DB A6 2F 20 88 66 98
                    :       85 34 2E EF D4 29 61 23 79 87 27 27 55 15 8D 21
         297 02   65:     INTEGER
                    :       00 F9 62 BD 22 4A C8 56 7A C3 17 EB CE CC 5F 42
                    :       E1 40 F5 A5 66 60 32 54 86 67 26 AD 7C 34 C2 FE
                    :       FE 8A F7 7F BE 79 53 5F C9 73 D9 47 8B 0F 89 A1
                    :       09 F1 27 16 FC F1 4B C3 A9 27 59 29 0D DA 9C AE
                    :       53
         364 02   65:     INTEGER
                    :       00 CF D1 4A 31 50 9A B4 BA 90 42 25 49 54 7C 20
                    :       54 2E CF E8 F1 35 DA 92 C2 A3 94 9D B7 B1 85 3F
                    :       13 D0 CA BC 77 D9 8A F3 32 83 59 93 E1 F0 11 1B
                    :       4C E5 A2 30 50 FE 1F B6 8D A5 B1 44 DA 4D 4B 11
                    :       09
         431 02   64:     INTEGER
                    :       46 53 3A C4 9D D4 0A D7 09 87 08 5F 43 B0 A5 5A
                    :       82 08 03 81 70 25 21 42 D9 79 C5 B8 5D E4 93 25
                    :       D2 A8 62 A4 A2 F0 08 F5 F5 2E 53 87 7A 75 34 2D
                    :       6A 8C BC 65 CD E1 B0 A6 55 CB 45 D1 7B 51 6D B3
         497 02   65:     INTEGER
                    :       00 81 CC 61 7F 9D AD 92 F5 F7 86 28 CD BD 43 ED
                    :       D9 46 87 BB 21 75 16 78 95 B3 1F EE C6 3D CD 50
                    :       91 6A D6 45 92 C1 C0 24 97 C7 2C 5A CE 42 68 1C
                    :       DA 11 8F 14 88 71 C0 92 FF B3 9E 9D B7 8F 91 34
                    :       29
         564 02   65:     INTEGER
                    :       00 88 7A 99 AC AA A9 D5 2B 6E E1 87 0A E8 D2 4C
                    :       04 8E A2 EA 00 3F 8D AF 9F 76 61 86 B0 1D 18 69
                    :       C8 64 22 D4 6B A3 A4 BB 52 B1 AC 38 DB 6B 5C 28
                    :       F0 78 73 3E 37 FD C8 54 72 C7 FD A9 EB C9 F2 45
                    :       96
                    :     }
                    :   }
    

    不幸的是,它不是正确编码的PKCS#8 PrivateKeyInfo。从索引22开始的序列是一个PKCS#1 PKCS1RSAPrivateKey,它应该被包装在一个八进制字符串中,以便对结构进行正确编码。

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

       0 30  631: SEQUENCE {
       4 02    1:   INTEGER 0
       7 30   13:   SEQUENCE {
       9 06    9:     OBJECT IDENTIFIER rsaEncryption (1 2 840 113549 1 1 1)
      20 05    0:     NULL
                :     }
      22 04  609:   OCTET STRING, encapsulates {
      26 30  605:       SEQUENCE {
      30 02    1:         INTEGER 0
      33 02  129:         INTEGER
                :           00 CA 72 B8 D1 B8 8E B9 39 C0 92 C1 4C 53 B4 F4
                :           38 48 3F C3 1C DC 6B BC BE 26 A3 B2 F7 7C 60 A8
                :           2C 0D 86 ED FC 2D D2 5C 99 B6 B6 71 A8 6D 2F 51
                :           25 FA 9C 42 FE 10 C1 2F 39 EA E8 FF 1A 78 BA 6B
                :           64 B8 39 34 3B F4 1C 45 06 C3 B9 98 DC 01 FF 41
                :           56 36 4F DD 35 69 A4 27 BB 5F FD DD 5C 73 BA 9A
                :           94 5A 4F 37 A9 48 3D 5B 89 EA EE BA 8D 02 6E D7
                :           6E D4 6F BC 7D 7A A4 41 4C 4D CA 08 05 20 66 A3
                :                   [ Another 1 bytes skipped ]
     165 02    3:         INTEGER 65537
     170 02  128:         INTEGER
                :           21 6A E2 7B 2B DD D3 51 67 2A 52 62 09 07 3B B0
                :           F6 AC 1F C6 E9 D3 96 EA 44 72 8D 1E 31 17 BB 6A
                :           DA 28 C5 AB F4 DC 5E 90 B9 0A 50 A4 9E B1 4A D1
                :           DC 16 63 30 91 0F 72 7E 3A FA 8E F1 8D B0 27 FD
                :           C2 BA B5 F8 FC 7C 46 C0 FD AD A7 39 7C 36 71 7A
                :           33 8B AD 0D 0C DA 50 B7 0E BF D8 64 7D 44 BD 64
                :           6F E2 51 B7 5E 2D 7B BA 02 DB A6 2F 20 88 66 98
                :           85 34 2E EF D4 29 61 23 79 87 27 27 55 15 8D 21
     301 02   65:         INTEGER
                :           00 F9 62 BD 22 4A C8 56 7A C3 17 EB CE CC 5F 42
                :           E1 40 F5 A5 66 60 32 54 86 67 26 AD 7C 34 C2 FE
                :           FE 8A F7 7F BE 79 53 5F C9 73 D9 47 8B 0F 89 A1
                :           09 F1 27 16 FC F1 4B C3 A9 27 59 29 0D DA 9C AE
                :           53
     368 02   65:         INTEGER
                :           00 CF D1 4A 31 50 9A B4 BA 90 42 25 49 54 7C 20
                :           54 2E CF E8 F1 35 DA 92 C2 A3 94 9D B7 B1 85 3F
                :           13 D0 CA BC 77 D9 8A F3 32 83 59 93 E1 F0 11 1B
                :           4C E5 A2 30 50 FE 1F B6 8D A5 B1 44 DA 4D 4B 11
                :           09
     435 02   64:         INTEGER
                :           46 53 3A C4 9D D4 0A D7 09 87 08 5F 43 B0 A5 5A
                :           82 08 03 81 70 25 21 42 D9 79 C5 B8 5D E4 93 25
                :           D2 A8 62 A4 A2 F0 08 F5 F5 2E 53 87 7A 75 34 2D
                :           6A 8C BC 65 CD E1 B0 A6 55 CB 45 D1 7B 51 6D B3
     501 02   65:         INTEGER
                :           00 81 CC 61 7F 9D AD 92 F5 F7 86 28 CD BD 43 ED
                :           D9 46 87 BB 21 75 16 78 95 B3 1F EE C6 3D CD 50
                :           91 6A D6 45 92 C1 C0 24 97 C7 2C 5A CE 42 68 1C
                :           DA 11 8F 14 88 71 C0 92 FF B3 9E 9D B7 8F 91 34
                :           29
     568 02   65:         INTEGER
                :           00 88 7A 99 AC AA A9 D5 2B 6E E1 87 0A E8 D2 4C
                :           04 8E A2 EA 00 3F 8D AF 9F 76 61 86 B0 1D 18 69
                :           C8 64 22 D4 6B A3 A4 BB 52 B1 AC 38 DB 6B 5C 28
                :           F0 78 73 3E 37 FD C8 54 72 C7 FD A9 EB C9 F2 45
                :           96
                :         }
                :       }
                :   }
    

    要修复您的文件,您可以使用ASN.1库(但我不知道Java有好的库),或者执行以下操作:

    检查您的数据是否以 30(*1)020100300D06092A864886F70D010101050030(*2) (*1) (*2) 将使用以下形式之一的长度编码

    • 长度<=0x7F: XX ,其中XX是长度
    • 0x80<=长度<=0xFF: 81XX ,其中XX是长度
    • 0x0100<=长度<=0xFFFF: 82XXXX ,其中XXXX是长度
    • 0x010000<=长度<=0xFFFFFF: 83XXXXXX

    如果键的长度都相同,则可以假定长度编码始终位于表单上 82XXXX

    把长度读进去 (*2) ,以字节为单位添加 30(*2) 到数字(这可能是4)和编码的长度如上所述(将最有可能是形式 82XXXX (*3) . 插入 04(*3) 30(*2) . 现在添加 04(*3) (*1) 然后重新编码(可能仍然适合) (*1) 用这个。

    我希望这是可以理解的,否则我建议阅读 A Layman's Guide to a Subset of ASN.1, BER, and DER

        2
  •  0
  •   Chochos    17 年前

    你试过用一些Bouncy Castle内部类打开钥匙吗?也许通过直接使用它们,而不是仅仅将BC定义为加密提供程序,您可以解析该文件。。。

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