[python] check if a file is open in Python

Using

try:
with open("path", "r") as file:#or just open

may cause some troubles when file is opened by some other processes (i.e. user opened it manually). You can solve your poblem using win32com library. Below code checks if any excel files are opened and if none of them matches the name of your particular one, openes a new one.

import win32com.client as win32
xl = win32.gencache.EnsureDispatch('Excel.Application')

my_workbook = "wb_name.xls"
xlPath="my_wb_path//" + my_workbook


if xl.Workbooks.Count > 0:
    # if none of opened workbooks matches the name, openes my_workbook 
    if not any(i.Name == my_workbook for i in xl.Workbooks): 
        xl.Workbooks.Open(Filename=xlPath)
        xl.Visible = True
#no workbooks found, opening
else:  
    xl.Workbooks.Open(Filename=xlPath)
    xl.Visible = True

'xl.Visible = True is not necessary, used just for convenience'

Hope this will help