DFM文件与XML文件互转

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在:DFM文件与XML文件互转中,用到的dfm文件必须为文本格式, 如果是二进制格式, 处理就会出错.

但是在处理中如何判断dfm是二进制文件, 而且再将二进制文件转为文本格式呢. ---

dfm文件二进制格式时, 其文件会加一个文件头, 其中前3个字节来标识其为二进制, 这三个字节分别为:$FF, $0A, $00. 因为这三个字节在文本类型的文件中是不可能存在的,所以可以判断这3个字节就可以了.

function IsBinDfm(const ADfmFileName: string): Boolean;Var  mBinStream:TMemoryStream;  mBuff : array [0..2] of byte;begin  mBinStream := TMemoryStream.Create;  try    mBinStream.LoadFromFile(ADfmFileName);    mBinStream.Read(mBuff, 3);    //前三字节: $FF, $0A, $00    if (mBuff[0] = $FF) and (mBuff[1] = $0A) and (mBuff[2]= $00) then      Result := True    else      Result := False;  finally    mBinStream.Free;  end;end;

判断出来后, 再将二进制转为文本格式就容易了.Delphi提供了ObjectResourceToText函数.写法如下:

procedure DfmBin2Txt(ADfmFileName: string);Var  inFileStream: TMemoryStream;  outFileStream: TFileStream;begin  inFileStream := TMemoryStream.Create;  inFileStream.LoadFromFile(ADfmFileName);  try    outFileStream := TFileStream.Create(ADfmFileName, fmCreate);    try      try        inFileStream.Seek(0, soFromBeginning);        ObjectResourceToText(inFileStream, outFileStream);      except        Raise Exception.Create('This dfm is bin, error on trans bin to txt.');      end;    finally      outFileStream.Free;    end;  finally    inFileStream.Free;  end;end;

至此,大功告成!