上传图片后,图片打开,高手们来帮我看看
procedure TForm1.Button3Click(Sender: TObject);
var
  response: TStringStream;
  MMPFDS: TIdMultiPartFormDataStream;
  idhttp1: TIDhttp;
begin
  if OpenDialog1.Execute then
  begin
    MMPFDS := TIdMultiPartFormDataStream.Create;
    response := TStringStream.Create('');
    idhttp1 := TIDhttp.Create(nil);
    MMPFDS.AddFile('file1', OpenDialog1.FileName, 'multipart/form-data');      //
    idhttp1.Request.ContentType := MMPFDS.RequestContentType;    MMPFDS.Position := 0;
    try
      idhttp1.Post('http://10.11.18.3/Upload/SoftUpload.asp', MMPFDS, response);
    finally
      showMessage(response.DataString);
      MMPFDS.Free;
      response.Free;
      idhttp1.Disconnect;
      idhttp1.Free;
    end;
  end;
end;
-----------------------------
ASP:
<%
bFileData=Request.BinaryRead(Request.TotalBytes)' 创建 Stream 对象
set ado_stream = Server.CreateObject("ADODB.Stream")' 打开Stream对象,把数据存入其中 
ado_stream.Type = 1 ' 1=adTypeBinary 
ado_stream.open 
ado_stream.Write bFileData' 文件存盘
ado_stream.SaveToFile Server.MapPath("a.jpg"),2 ' 2=adSaveCreateOverWrite 
ado_stream.close' 销毁对象 
set ado_stream = Nothing 
' 向浏览器返回信息if -2147221005 <> Err then
Response.Write "ok"
else
Response.Write "Err" & chr(13)
Response.Write Err.Number & ":" &  Err.Description
end if
%>