Convertir l'image BMP au format GRF C#/VB.NET (à utiliser dans l'imprimante ZPL)

Question

J'utilise le code suivant pour convertir l'image BMP au format GRF.

Public Shared Function CreateGrf(filename As String, imagename As String) As String 
    Dim bmp As Bitmap = Nothing 
    Dim imgData As BitmapData = Nothing 
    Dim pixels As Byte() 
    Dim x As Integer, y As Integer, width As Integer 
    Dim sb As StringBuilder 
    Dim ptr As IntPtr 

    Try 
     bmp = New Bitmap(filename) 
     imgData = bmp.LockBits(New System.Drawing.Rectangle(0, 0, bmp.Width, bmp.Height), ImageLockMode.ReadWrite, bmp.PixelFormat) 
     width = (bmp.Width + 7) \ 8 
     pixels = New Byte(width - 1) {} 
     sb = New StringBuilder(width * bmp.Height * 2) 
     sb.Append(Environment.NewLine) 
     ptr = imgData.Scan0 

     For y = 0 To bmp.Height - 1 
      Marshal.Copy(ptr, pixels, 0, width) 
      For x = 0 To width - 1 
       sb.AppendFormat("{0:X2}", CByte(Not pixels(x))) 
      Next 
      sb.Append(Environment.NewLine) 
      ptr = ptr.ToInt64() + imgData.Stride 
     Next 
    Finally 
     If bmp IsNot Nothing Then 
      If imgData IsNot Nothing Then 
       bmp.UnlockBits(imgData) 
      End If 
      bmp.Dispose() 
     End If 
    End Try 
    Return [String].Format("~DG{0},{1},{2},", imagename, width * y, width) + sb.ToString() 
End Function 

Cependant, il y a une ligne verticale supplémentaire tracée à la fin du fichier converti GRF même si il n'y a pas une telle ligne dans le fichier BMP. A part ça la taille et tout est OK. Il semble que le dernier pixel (valeur hexadécimale) de chaque ligne n'est pas correct dans le fichier GRF.

fichier original BMP.

transformé GRF FIle

Answer 1

Public Function ConvertBmp2Grf(fileName As String, imageName As String) As Boolean 
    Dim TI As String 
    Dim i As Short 
    Dim WID As Object 
    Dim high As Object 
    Dim TEM As Short, BMPL As Short, EFG As Short, n2 As String, LON As String 
    Dim header_name As String, a As String, j As Short, COUN As Short, BASE1 As Short 

    Dim L As String, TOT As String 
    Dim N As Object 
    Dim TOT1 As Integer 
    Dim LL As Byte 

    FileOpen(1, fileName, OpenMode.Binary, , , 1) ' OPEN BMP FILE TO READ 
    FileGet(1, LL, 1) 
    TI = Convert.ToString(Chr(LL)) 
    FileGet(1, LL, 2) 
    TI += Convert.ToString(Chr(LL)) 

    If TI <> "BM" Then 
     FileClose() 
     Return False 
    End If 

    i = 17 
    FileGet(1, LL, i + 1) 
    N = LL * 256 
    FileGet(1, LL, i) 
    N = (N + LL) * 256 

    FileGet(1, LL, i + 3) 
    N = (N + LL) * 256 
    FileGet(1, LL, i + 2) 
    N += LL 
    WID = N 
    i = 21 
    FileGet(1, LL, i + 1) 
    N = LL * 256 
    FileGet(1, LL, i) 
    N = (N + LL) * 256 
    FileGet(1, LL, i + 3) 
    N = (N + LL) * 256 
    FileGet(1, LL, i + 2) 
    N += LL 
    high = N 
    FileGet(1, LL, 27) 
    N = LL 
    FileGet(1, LL, 29) 

    If N <> 1 Or LL <> 1 Then 
     'BMP has too many colors, only support monochrome images 
     FileClose(1) 
     Return False 
    End If 

    TEM = Int(WID/8) 
    If (WID Mod 8) <> 0 Then 
     TEM += 1 
    End If 
    BMPL = TEM 

    If (BMPL Mod 4) <> 0 Then 
     BMPL += (4 - (BMPL Mod 4)) 
     EFG = 1 
    End If 

    n2 = fileName.Substring(0, fileName.LastIndexOf("\", StringComparison.Ordinal) + 1) + imageName + ".GRF" 

    FileOpen(2, n2, OpenMode.Output) 'OPEN GRF TO OUTPUT 
    TOT1 = TEM * high : TOT = Mid(Str(TOT1), 2) 
    If Len(TOT) < 5 Then 
     TOT = Strings.Left("00000", 5 - Len(TOT)) + TOT 
    End If 

    LON = Mid(Str(TEM), 2) 

    If Len(LON) < 3 Then 
     LON = Strings.Left("000", 3 - Len(LON)) + LON 
    End If 

    header_name = imageName 
    PrintLine(2, "~DG" & header_name & "," & TOT & "," & LON & ",") 

    For i = high To 1 Step -1 
     a = "" 
     For j = 1 To TEM 
      COUN = 62 + (i - 1) * BMPL + j 
      FileGet(1, LL, COUN) 
      L = LL 

      If j = TEM And (EFG = 1 Or (WID Mod 8) <> 0) Then 
       BASE1 = 2^((TEM * 8 - WID) Mod 8) 
       L = Int(L/BASE1) * BASE1 + BASE1 - 1 
      End If 
      L = Not L 
      a += Right(Hex(L), 2) 
     Next j 
     PrintLine(2, a) 
    Next i 
    FileClose() 

    Return True 

End Function 

Answer 2

Marshal.Copy (ptr, pixels, 0, largeur)

Le Bitmap est pas aligné octet. Donc, dans ce cas, lorsque vous copiez les données, les bits restants sont remplis en noir. La bitmap a une largeur de 154 octets, ce qui crée 19 octets complets et 2 pixels restants. Les 6 pixels restants sont donc noirs. En fin de compte, vous devez utiliser des bitmaps avec des largeurs divisibles par huit ou vous assurer que la fin de la copie de données de la bitmap vers les pixels (x) compte pour les octets restants.