there are two options: 1.Place a shortcut in the startup item of the programs menu 2.Place a registry key in the HKey -> currentuser -> software -> windows -> current version -> run you can find out how to do these through google1 Posted by The ANZAC : "WWBD, What Would Buffy Do?" : "I don't know man, she's stronger than me"

You can put it in the 'run' key of the registry, in the 'startup' group of the start menu, or in the old, text config files that windows still uses. Christian Graus - Microsoft MVP - C++ "I am working on a project that will convert a FORTRAN code to corresponding C++ code.I am not aware of FORTRAN syntax" ( spotted in the C++/CLI forum )

Hi all, Sorry to be late. Dave Kreskowiak wrote: So what does this mean? I meant, I can open any picture i stored in sql database in a picturebox. I stored picture as bytes. Opening of first and second picture (randomly from database) work smoothly. But when i try to open third image then it will show that error. I used following function to retrieve image from collection of bytes. Public Function PictureFromBits(abPic() As Byte) As IPicture ' abPic() as a byte array Dim nLow As Long Dim cbMem As Long Dim hMem As Long Dim lpMem As Long Dim IID_IPicture As GUID Dim istm As stdole.IUnknown Dim ipic As IPicture On Error GoTo Out GlobalFree (hMem) nLow = LBound(abPic) On Error GoTo 0 cbMem = (UBound(abPic) - nLow) + 1 hMem = GlobalAlloc(GMEM_MOVEABLE, cbMem) If hMem Then lpMem = GlobalLock(hMem) If lpMem Then MoveMemory ByVal lpMem, abPic(nLow), cbMem Call GlobalUnlock(hMem) If (CreateStreamOnHGlobal(hMem, CTrue, istm) = S_OK) Then If (CLSIDFromString(StrPtr(sIID_IPicture), IID_IPicture) = S_OK) Then Call OleLoadPicture(ByVal ObjPtr(istm), cbMem, CFalse, IID_IPicture, PictureFromBits) End If End If End If End If GlobalFree (hMem) Exit Function Out: MsgBox "Out" End Function I think this is a problem regarding memory. Please help. Dave Kreskowiak wrote: How big are these images, in bytes? I opened images with almost all sizes. eg: 57060 Bytes. Thanking for your interest.

The problem is that you're using VB

For Downloding... 'Declare on Gen.DECLARATION Private Declare Function ShellExecute Lib "shell32.dll" Alias "ShellExecuteA" (ByVal hwnd As Long, ByVal lpOperation As String, ByVal lpFile As String, ByVal lpParameters As String, ByVal lpDirectory As String, ByVal nShowCmd As Long) As Long 'This line is also declared on Declaration with both,uploading and downloading Dim strStream As ADODB.Stream Dim RstRpt As New ADODB.Recordset Set RstRpt = ModFncRecordSetReturnRst("Select FILECONTENT from DMSDOCCONTENTTRN" _ & " Where DOCUMENTCODE ='" & txtcode.text & "'") If RstRpt.RecordCount > 0 Then Set strStream = New ADODB.Stream strStream.Type = adTypeBinary strStream.Open strStream.Write RstRpt.Fields("FILECONTENT").Value DestinationFile = "C:\" & txtcode.text strStream.SaveToFile (DestinationFile), adSaveCreateOverWrite 'FileCopy SourceFile, DestinationFile SetAttr DestinationFile, vbNormal Call ShellExecute(hwnd, "Open", DestinationFile, "", "C:\", 1) 'it will save it in C:\ and Open it from the C:\ Path 'In case of Any Trouble Ask me. Take Care. Bye. Regards; Ali

i don't know if i completely understand you BUT if you write your html to any file with File.WriteAllText("yourfile.ddd", strhtml, System.Text.Encoding.UTF8) then with a simple webbrowser control you can view it WebBrowser1.Refresh() WebBrowser1.Navigate("yourfile.ddd") hope that helps isidoros

Well, There are morethan 25 languages in India, not only Hindi, and may be its the same in one of these languages ;) just jocking Vilsad P P MCTS (Windows Applications) .Net 2.0