java如何从奥地利ecard将ResponseADU解码为XML？

Question

我试着从一张奥地利电子贺卡上读取信息，以获得名字和姓氏

目前有效的方法是：访问卡，发送APDU命令，并以字节数组的形式获取信息

如何将接收到的字节数组转换为XML以提取所需的数据

以下是代码：

import java.util.List;

import javax.smartcardio.Card;
import javax.smartcardio.CardChannel;
import javax.smartcardio.CardException;
import javax.smartcardio.CardTerminal;
import javax.smartcardio.CommandAPDU;
import javax.smartcardio.ResponseAPDU;
import javax.smartcardio.TerminalFactory;

public class Main2 {
    public static void main(String[] args) {
        TerminalFactory factory = TerminalFactory.getDefault();
        List<CardTerminal> terminals;
        try {
            terminals = factory.terminals().list();
            CardTerminal terminal = terminals.get(0);
            Card card = terminal.connect("*");
            CardChannel channel = card.getBasicChannel();
            // Select the MF
            byte[] aid = { (byte) 0xD0, 0x40, 0x00, 0x00, 0x17, 0x01, 0x01, 0x01 };
            ResponseAPDU resp = channel.transmit(new CommandAPDU(0x00, 0xA4, 0x04, 0x00, aid));
            System.out.println("Response: " + resp.toString());
            // Select the Personaladata-file
            byte[] aid2 = { (byte) 0xEF, 0x01 };
            resp = channel.transmit(new CommandAPDU(0x00, 0xA4, 0x02, 0x04, aid2));
            System.out.println("Response: " + resp.toString());
            // Get the data from the file
            resp = channel.transmit(new CommandAPDU(0x00, 0xB0, 0x00, 0x00, 0xFF));
            System.out.println("Response: " + resp.toString());
            System.out.println("Response String: " + new String(resp.getData()));
            card.disconnect(false);
        } catch (CardException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
}

Answer 1

感谢您的提示，下面是将DER字节数组解码为字符串的代码

ASN1InputStream input = new ASN1InputStream(resp.getData());
            ASN1Primitive p;
            try {
                while ((p = input.readObject()) != null) {
                    // System.out.println("DEBUG: " + ASN1Dump.dumpAsString(p));
                    // Sozialversicherungsnummer
                    ASN1Sequence asn1 = ASN1Sequence.getInstance(p);
                    ASN1Sequence seq = DLSequence.getInstance(asn1.getObjectAt(0));
                    ASN1Set svn = DLSet.getInstance(seq.getObjectAt(1));
                    DERNumericString svnObject = DERNumericString.getInstance(svn.getObjectAt(0));
                    System.out.println("SVN: " + svnObject.getString());

                    // Vorname
                    seq = DLSequence.getInstance(asn1.getObjectAt(2));
                    svn = DLSet.getInstance(seq.getObjectAt(1));
                    DERUTF8String stringObject = DERUTF8String.getInstance(svn.getObjectAt(0));
                    System.out.println("Vorname: " + stringObject.getString());

                    // Nachname
                    seq = DLSequence.getInstance(asn1.getObjectAt(3));
                    svn = DLSet.getInstance(seq.getObjectAt(1));
                    stringObject = DERUTF8String.getInstance(svn.getObjectAt(0));
                    System.out.println("Vorname: " + stringObject.getString());

                    // Geschlecht
                    seq = DLSequence.getInstance(asn1.getObjectAt(5));
                    svn = DLSet.getInstance(seq.getObjectAt(1));
                    DERPrintableString charObject = DERPrintableString.getInstance(svn.getObjectAt(0));
                    System.out.println("Geschlecht: " + charObject.getString());
                }
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

Answer 2

我不知道如何将数据转换成XML结构（以及根据什么模式）。然而，我从SV卡收到的字节数组看起来像ASN。1 DER编码的TLV结构：

30 xxxx
SEQUENCE
    30 18
    SEQUENCE
        06 08
        OBJECT IDENTIFIER
            2A28000A01040101
            => OID 1.2.40.0.10.1.4.1.1 (SV number)
        31 0C
        SET
            12 0A
            NumericString
                nnnnnnnnddddmmmmyyyy
                => SV number: NNNN DDMMYY
    30 0F
    SEQUENCE
        06 08
        OBJECT IDENTIFIER
            2A28000A01040103
            => OID 1.2.40.0.10.1.4.1.3 (Card sequence number)
        31 03
        SET
            02 01
            INTEGER
                xx
                => Card sequence number: xx
    30 xx
    SEQUENCE
        [...]
    30 xx
        SEQUENCE
        06 03
        OBJECT IDENTIFIER
            55042A
            => OID 2.5.4.42 ({joint-iso-itu-t(2) ds(5) attributeType(4) givenName(42)})
        31 xx
        SET
            0C xx
            UTF8String
                4D69636861656C
                => Given name: "Michael"
    30 xx
    SEQUENCE
        06 03
        OBJECT IDENTIFIER
            550404
            => OID 2.5.4.4 ({joint-iso-itu-t(2) ds(5) attributeType(4) surname(4)})
        31 xx
        SET
            0C xx
            UTF8String
                526F6C616E64
                => Surname: "Roland"
    30 xx
    SEQUENCE
        [...]
    30 1D
    SEQUENCE
        06 08
        OBJECT IDENTIFIER
            2B06010505070901
            => OID 1.3.6.1.5.5.7.9.1 ({iso(1) identified-organization(3) dod(6) internet(1) security(5) mechanisms(5) pkix(7) pda(9) dateOfBirth(1)})
        31 11
        SET
            18 0F
            GeneralizedTime
                yyyyyyyymmmmdddd3132303030305A
                => Date of birth: YYYY-MM-DD 12:00:00Z
    30 0F
    SEQUENCE
        06 08
        OBJECT IDENTIFIER
            2B06010505070903
            => OID 1.3.6.1.5.5.7.9.3 ({iso(1) identified-organization(3) dod(6) internet(1) security(5) mechanisms(5) pkix(7) pda(9) gender(3)})
        31 03
        SET
            13 01
            PrintableString
                4D
                => Gender: M (male)

所以这似乎遵循了下面的ASN。1表示法：

SVPersonGrunddaten ::= SEQUENCE OF Attribute
Attribute ::= SEQUENCE {
    attributeName OBJECT IDENTIFIER,
    attributeValue SET OF AttributeType }
AttributeType ::= CHOICE {
    numericString NumericString,
    integer INTEGER,
    utf8String UTF8String,
    time GeneralizedTime,
    printableString PrintableString }

其中，给定名称和姓氏的属性为

givenName Attribute ::= {
    attributeName 2.5.4.42,
    attributeValue { utf8String "Given Name" }
}
surname Attribute ::= {
    attributeName 2.5.4.4,
    attributeValue { utf8String "Surname" }
}

因此，为了获得给定的名称和姓氏，需要解析TLV结构，搜索这两个元素的OID，并将相关值解码为UTF8字符串

请注意，简单地假设字段位于准确的位置似乎不是一个好主意。例如，在名字字段前面有一个字段30 xx ...（类型为Attribute的字段），似乎只有在卡片上打印了学术/专业头衔（如“Dr.”在我的情况下）时才会出现。同样，学术后缀还有另一个可选字段（如“M.Sc.”），只有在卡片上打印了这样的后缀时才会出现。虽然我的卡片上所有其他字段的顺序都是一样的，但我不确定这是否是必需的