I’m using the following code to open and display a workbook within Excel:
import win32com.client as win32 excel = win32.gencache.EnsureDispatch('Excel.Application') wb = excel.Workbooks.Open('my_sheet.xlsm') ws = wb.Worksheets('blaaaa') excel.Visible = True
When the File ‘my_sheet.xlsm’ is already opened, Excel asks me whether I want to re-open it without saving.
How could I check in advance whether the workbook is already opened up and in case it is, just bring it to the front?
EDIT: Found out by now:
if excel.Workbooks.Count > 0: for i in range(1, excel.Workbooks.Count+1): if excel.Workbooks.Item(i).Name is 'my_sheet.xlsm': wb = excel.Workbooks.Item(i) break
And one more question: My worksheet contains some headers where I enabled some filtering. So when the filters are set, and when I open the Workbook from Python, it sometimes asks me to enter a unique name to save the filter. Why is that? This is the dialogue:
EDIT Ok here it says (in german language) that the latter problem is a known bug in 2007 and 2010 files: https://social.msdn.microsoft.com/Forums/de-DE/3dce9f06-2262-4e22-a8ff-5c0d83166e73/excel-api-interne-namen and it seems to exist if you open Excel-Files programmatically. Don’t know whether there is a workaround.
While you found a solution, consider using try/except/finally block. Currently your code will leave the Excel.exe process running in background (check Task Manager if using Windows) even if you close the visible worksheet. As an aside, in Python or any other language like VBA, any external API such as this COM interface should always be released cleanly during application code.
Below solution uses a defined function,
openWorkbook() to go two potential routes: 1) first attempts to relaunch specified workbook assuming it is opened and 2) if it is not currently opened launches a new workbook object of that location. The last nested
try/except is used in case the
Workbooks.Open() method fails such as incorrect file name.
import win32com.client as win32 def openWorkbook(xlapp, xlfile): try: xlwb = xlapp.Workbooks(xlfile) except Exception as e: try: xlwb = xlapp.Workbooks.Open(xlfile) except Exception as e: print(e) xlwb = None return(xlwb) try: excel = win32.gencache.EnsureDispatch('Excel.Application') wb = openWorkbook(excel, 'my_sheet.xlsm') ws = wb.Worksheets('blaaaa') excel.Visible = True except Exception as e: print(e) finally: # RELEASES RESOURCES ws = None wb = None excel = None