Konvertieren von Byte-Array in ein Bild

Question

Ich versuche, ein Bild von einem Smartcard-Leser zu bekommen. Es kam mit einer Beispiel-C# -Lösung, die Karteninformationen und das Portrait des Karteninhabers anzeigt. Leider taucht das Porträt nie auf. Es ist eigentlich mit folgendem Inhalt in eine Datei geschrieben:

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    ftypjpx  jpx jp2 jpxb 1rreqÿ ÿ € +@     ‹jp2h ihdr > î Scolr HLino mntrRGB XYZ Î  1 acspMSFT IEC sRGB    öÖ  Ó-HP             cprt P 3desc „ lwtpt ð bkpt  rXYZ  gXYZ , bXYZ @ dmnd T pdmdd Ä ˆvued L †view Ô $lumi ø meas  $tech 0 rTRC < gTRC < bTRC < text Copyright (c) 1998 Hewlett-Packard Company desc  sRGB IEC61966-2.1   sRGB IEC61966-2.1             XYZ  óQ ÌXYZ     XYZ  o¢ 8õ XYZ  b™ ·… ÚXYZ  $  „ ¶Ïdesc  IEC http://www.iec.ch   IEC http://www.iec.ch            desc  .IEC 61966-2.1 Default RGB colour space - sRGB   .IEC 61966-2.1 Default RGB colour space - sRGB      desc  ,Reference Viewing Condition in IEC61966-2.1   ,Reference Viewing Condition in IEC61966-2.1       view  ¤þ _. Ï íÌ \ž XYZ  L V P Wçmeas       sig  CRT curv   
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3F33¸3ñ4+4e4ž4Ø55M5‡5Â5ý676r6®6é7$7`7œ7×88P8Œ8È99B99¼9ù:6:t:²:ï;-;k;ª;è<'<e<¤<ã="=a=¡=à> >`> >à?!?a?¢?â@#@d@¦@çA)AjA¬AîB0BrBµB÷C:C}CÀDDGDŠDÎEEUEšEÞF"FgF«FðG5G{GÀHHKH‘H×IIcI©IðJ7J}JÄKKSKšKâL*LrLºMMJM“MÜN%NnN·O OIO“OÝP'PqP»QQPQ›QæR1R|RÇSS_SªSöTBTTÛU(UuUÂVV\V©V÷WDW’WàX/X}XËYYiY¸ZZVZ¦Zõ[E[•[å\5\†\Ö]']x]É^^l^½__a_³``W`ª`üaOa¢aõbIbœbðcCc—cëd@d”dée=e’eçf=f’fèg=g“géh?h–hìiCišiñjHjŸj÷kOk§kÿlWl¯mm`m¹nnknÄooxoÑp+p†pàq:q•qðrKr¦ss]s¸ttptÌu(u…uáv>v›vøwVw³xxnxÌy*y‰yçzFz¥{{c{Â|!||á}A}¡~~b~Â#„å€G€¨ 
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æ–çç©è2è¼éFéÐê[êåëpëûì†ííœî(î´ï@ïÌðXðåñrñÿòŒóó§ô4ôÂõPõÞömöû÷Šøø¨ù8ùÇúWúçûwüü˜ý)ýºþKþÜÿmÿÿ res resc€ € jpch jplh "®jp2cÿOÿQ/ î >      ÿR ÿ\ PXX`XX`XX`XXXPPXÿd Kakadu-v6.3.1ÿd Kdu-Layer-Info: log_2{Delta-D(MSE)/[2^16*Delta-L(bytes)]}, L(bytes) 
-45.0, 6.0e+002 
-46.3, 7.0e+002 
-47.2, 8.2e+002 
-48.5, 9.9e+002 
-49.4, 1.2e+003 
-50.2, 1.6e+003 
-51.5, 2.0e+003 
-52.6, 2.6e+003 
-53.8, 3.5e+003 
-55.2, 4.8e+003 
-56.2, 6.5e+003 
-57.8, 9.0e+003 
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Es hat einige Zeile mit folgendem Text:

Copyright (c) 1998 Hewlett-Packard Company desc  sRGB IEC61966-2.1   sRGB IEC61966-2.1 

Der Code:

cmdSelect = new APDUCmd(0x00, 0xA4, 0x00, 0x00, new byte[] { 0x3F, 0x00 }, 0x00); 
cmdResponse = cardOri.Transmit(cmdSelect); 
cmdSelect = new APDUCmd(0x00, 0xA4, 0x04, 0x00, new byte[] { 0x49, 0x44 }, 0x00); 
cmdResponse = cardOri.Transmit(cmdSelect); 
cmdSelect = new APDUCmd(0x00, 0xA4, 0x02, 0x00, new byte[] { 0x01, 0x01 }, 0x00); 
cmdResponse = cardOri.Transmit(cmdSelect); 
cmdSelect = new APDUCmd(0x00, 0xA4, 0x02, 0x00, new byte[] { 0x01, 0x02 }, 0x00); 
cmdResponse = cardOri.Transmit(cmdSelect); 
List<string> picData = new List<string>(); 
List<byte> picData1 = new List<byte>(); 
List<byte> picData2 = new List<byte>(); 
int j = 0x05; 
int i = 0x00; 
string sst = string.Empty; 
int p1_first = 0x00; 
int p2_first = 0x00; 
int p1_end = 0x00; 
int p2_end = 0x00; 
int count = 0; 

while (i <= 0x30) 
{ 

    count++; 
    try 
    { 

     cmdSelect = new APDUCmd(0x00, 0xB0, (byte)i, (byte)j, null, 0xFF); 
     cmdResponse = cardOri.Transmit(cmdSelect); 
     if (cmdResponse.Data != null) 
     { 
      picData1.AddRange(cmdResponse.Data); 
     } 
     else 
     { 
      p1_first = i - 1; 
      p2_first = j + 1; 

      byte[] ddd = null; 
      while (true) 
      { 

       j--; 
       cmdSelect = new APDUCmd(0x00, 0xB0, (byte)i, (byte)j, null, 0xFF); 
       cmdResponse = cardOri.Transmit(cmdSelect); 

       if (cmdResponse.Data != null) 
       { 
        ddd = cmdResponse.Data; 
        picData2.AddRange(ddd); 
        p1_end = i; 
        p2_end = j; 
        break; 
       } 
       else 
       { 
        // break; 
       } 
       if (j <= 0x00) 
       { 
        i--; 
        j = 0xFF; 
       } 
      } 
      break; 
     } 

    } 
    catch (Exception ex) 
    { 
     MessageBox.Show(ex.Message, "Error", MessageBoxButtons.OK, MessageBoxIcon.Error); 
     DisConnect(); 
     break; 
    } 
    j = j + 0xFF; 
    if (j > 0xFF) 
    { 
     i++; 
     j = j - 0x100; 
    } 

} 
int len = 0; 
if (p1_first == p1_end) 
{ 
    if (p2_first > p2_end) 
    { 
     len = 255 - (p2_first - p2_end); 
    } 
    else 
    { 
     len = 255 - (p2_end - p2_first); 
    } 
} 
else 
{ 
    if (p2_first > p2_end) 
    { 
     len = p2_first - p2_end - 1; 
    } 
    else 
    { 
     len = p2_end - p2_first - 1; 
    } 

} 


int k = 0; 
for (int l = len; l < 255; l++) 
{ 
    picData1.Add(picData2[l]); 
    k++; 
} 

File.WriteAllBytes("pic.txt", picData1); 

Methoden Ich habe versucht:

public Image byteArrayToImage(byte[] byteArrayIn) 
    { 
     MemoryStream ms = new MemoryStream(byteArrayIn); 
     Image returnImage = Image.FromStream(ms); 
     return returnImage; 
    } 

public bool ByteArrayToFile(string _FileName, byte[] _ByteArray) 
    { 
     try 
     { 
      System.IO.FileStream _FileStream = new System.IO.FileStream(_FileName, System.IO.FileMode.Create, System.IO.FileAccess.Write); 
      _FileStream.Write(_ByteArray, 0, _ByteArray.Length); 
      _FileStream.Close(); 
      return true; 
     } 
     catch (Exception _Exception) 
     { 
      Console.WriteLine("Exception caught in process: {0}", _Exception.ToString()); 
     } 
     return false; 
    } 

Wie sollte ich konvergieren t das Byte arrey in ein Bild?

Answer 1

Wenn Ihre Frage richtig verstanden möchten Sie Bild aus dem Kartenleser zu Ihrem PC erhalten dann diese ausprobieren

byte[] ArrayOfBytesOfYourImage= System.IO.File.ReadAllBytes(Path); //Path your image file in sd card 
System.IO.File.WriteAllBytes(@"Your path you want to save with extenction for example C:\Foo.jpeg", ArrayOfBytesOfYourImage); 

wenn Sie Bild nicht, sondern haben nur Bytes. Dann brauchst du keine erste Zeile.