在Delphi10.4中,我成功地使用以下代码将有效的TPicture base64 64保存到INI文件中:
procedure TForm1.SavePictureToIniFile(const APicture: TPicture);
// https://stackoverflow.com/questions/63216011/tinifile-writebinarystream-creates-exception
var
LInput: TMemoryStream;
MyIni: TMemIniFile;
Base64Enc: TBase64Encoding;
ThisFile: string;
begin
if FileSaveDialog1.Execute then
ThisFile := FileSaveDialog1.FileName
else EXIT;
//CodeSite.Send('TForm1.btnSaveToIniClick: VOR Speichern');
LInput := TMemoryStream.Create;
try
APicture.SaveToStream(LInput);
LInput.Position := 0;
MyIni := TMemIniFile.Create(ThisFile);
try
Base64Enc := TBase64Encoding.Create(Integer.MaxValue, '');
try
MyIni.WriteString('Custom', 'IMG', Base64Enc.EncodeBytesToString(LInput.Memory, LInput.Size));
finally
Base64Enc.Free;
end;
MyIni.UpdateFile;
finally
MyIni.Free;
end;
finally
LInput.Free;
end;
//CodeSite.Send('TForm1.btnSaveToIniClick: NACH Speichern'); // 0,024 Sek.
end;现在,我想逆转这个过程,即将数据从INI文件加载回TPicture
procedure TForm1.btnLoadFromIniClick(Sender: TObject);
var
LInput: TMemoryStream;
LOutput: TMemoryStream;
ThisFile: string;
MyIni: TMemIniFile;
Base64Enc: TBase64Encoding;
ThisEncodedString: string;
ThisPicture: TPicture;
begin
if FileOpenDialog1.Execute then
ThisFile := FileOpenDialog1.FileName
else EXIT;
MyIni := TMemIniFile.Create(ThisFile);
try
Base64Enc := TBase64Encoding.Create(Integer.MaxValue, '');
try
(*ThisEncodedString := MyIni.ReadString('Custom', 'IMG', '');
Base64Enc.Decode(ThisEncodedString); // And now???*)
LInput := TMemoryStream.Create;
LOutput := TMemoryStream.Create;
try
MyIni.ReadBinaryStream('Custom', 'IMG', LInput);
MyIni.UpdateFile;
LInput.Position := 0;
Base64Enc.Decode(LInput, LOutput);
LOutput.Position := 0;
ThisPicture := TPicture.Create;
try
ThisPicture.LoadFromStream(LOutput);
CodeSite.Send('TForm1.btnLoadFromIniClick: ThisPicture', ThisPicture); // AV!
finally
ThisPicture.Free;
end;
finally
LOutput.Free;
LInput.Free;
end;
finally
Base64Enc.Free;
end;
finally
MyIni.Free;
end;
end;但是,当用CodeSite.Send发送图片时,会创建一个AV!(用TPicture发送CodeSite.Send通常是有效的,在这种情况下,AV显然意味着图片已经损坏)。
那么,如何将数据从INI文件加载回TPicture
发布于 2020-08-03 09:47:18
这与原来的问题实质上是同一个问题。
INI文件的数据是二进制图像的Base64表示,即字符串。因此,您需要读取这个Base64字符串,并使用Base64Enc将其转换为二进制blob。
但是,您的代码使用ReadBinaryStream方法,它不将文本视为Base64字符串,而是作为十六进制字节序列,并将其作为二进制blob返回,然后将其交给Base64Enc。
取而代之的是这样做:
var
ImgData: TBytes;
begin
MyIni := TMemIniFile.Create('D:\img.ini');
try
Base64Enc := TBase64Encoding.Create(Integer.MaxValue, '');
try
LInput := TMemoryStream.Create;
try
ImgData := Base64Enc.DecodeStringToBytes(MyIni.ReadString('Custom', 'IMG', ''));
LInput.WriteData(ImgData, Length(ImgData));
LInput.Position := 0;
ThisPicture := TPicture.Create;
try
ThisPicture.LoadFromStream(LInput);
// Use ThisPicture
finally
ThisPicture.Free;
end;
finally
LInput.Free;
end;
finally
Base64Enc.Free;
end;
finally
MyIni.Free;
end;你能意识到这一点的一种方法是这样想:
我该怎么编码?好吧,我知道
Base64Enc.EncodeBytesToStringMyIni.WriteString所以,为了解码,我用相反的顺序做相反的步骤:
MyIni.ReadStringBase64Enc.DecodeStringToBytes摆脱不必要的拷贝
在注释中,Remy Lebeau正确地指出,上面的代码对二进制图像数据执行不必要的内存副本。虽然这不太可能是一个问题(甚至是可测量的!)实际上,考虑到我们从INI文件中的Base64 64编码字段读取图像,它仍然是浪费和丑陋的。
通过将TMemoryStream替换为TBytesStream (TMemoryStream的后代),我们可以将Base64数据直接解码到流中:
var
ImgStream: TBytesStream;
begin
MyIni := TMemIniFile.Create('D:\img.ini');
try
Base64Enc := TBase64Encoding.Create(Integer.MaxValue, '');
try
ImgStream := TBytesStream.Create(Base64Enc.DecodeStringToBytes(MyIni.ReadString('Custom', 'IMG', '')));
try
ThisPicture := TPicture.Create;
try
ThisPicture.LoadFromStream(ImgStream);
// Use ThisPicture
finally
ThisPicture.Free;
end;
finally
ImgStream.Free;
end;
finally
Base64Enc.Free;
end;
finally
MyIni.Free;
end;https://stackoverflow.com/questions/63226950
复制相似问题