Hi, As you know, I can use this to open an Excel file: """ import win32com.client
doc = win32com.client.Dispatch("Excel.Application") doc.Workbooks.Open(excelFile, ReadOnly=True) """ But the problem is when I only pass the filename to the Open() method, and of course the file is in the current directory, it will throw an exception that the specified file can't be found. When I use os.path.abspath(excelFile) instead, it works. But I do know that somebody can only pass a filename, and no exception is raised. Now I wonder is that because the "somebody" happen to put the file on the default location of the Open() method? If so, does any one know the default location? Thanks for your consideration. Regards, Johnny -- http://mail.python.org/mailman/listinfo/python-list