In Vista the user is getting Run Time Error '75' Path/File access error with the following procedure (primitive procedure).
Private Sub command1_Click()
Dim FILENUM
FILENUM = FreeFile
Open "C:\AA.txt" For Output As FILENUM
Print #FILENUM, "test..."
Close FILENUM
MsgBox ("C:\AA.txt was successfully created.")
End Sub
This procedure included in VB EXE file.
Question: In Vista Open For Output doesn't work and the VB code must be rewritten or maybe there is a setting in Vista allowing to create in VB code new file (at least text type)?
Thank you,
Alex Borodulin
http://www.nyacad.com
