Specify filename in python script to open Excel workbook

Question

this is a really dumb question, but Im trying to learn python and I got stuck on example with reading Excel files using xlrd. I found this script online but I cant figure out where am I supposed to fill my filename to get it open.

from future import print_function
from os.path import join, dirname, abspath, isfile
from collections import Counter
import xlrd
from xlrd.sheet import ctype_text   

def get_excel_sheet_object(fname, idx=0):
    if not isfile(fname):
        print ('File doesn't exist: ', fname)
            # Open the workbook and 1st sheet
    xl_workbook = xlrd.open_workbook(fname)
    xl_sheet = xl_workbook.sheet_by_index(0)
    print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)

return xl_sheetsdf

Antoine Gaia · Accepted Answer · 2016-12-14 09:28:48Z

0

You can enter the filename :

When you call the function

get_excel_sheet_object("myfile.xlsx")
```
       OR
```
fname = "myfile.xlsx"

get_excel_sheet_object(fname)

Raw in your program :

def get_excel_sheet_object(idx=0):

      fname = "myfile.xlsx"

      if not isfile(fname):
         print ("File doesn't exist: ", fname)
      # Open the workbook and 1st sheet
      xl_workbook = xlrd.open_workbook(fname)
      xl_sheet = xl_workbook.sheet_by_index(0)
      print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)

      return xl_sheet

answered Dec 14, 2016 at 9:28

Antoine Gaia

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3 Comments

k_mishap Over a year ago

Thanks but still doesnt work. I have the excel file in the same folder, do I need to specify the full path nevertheles?

k_mishap Over a year ago

well, it said the syntax is wrong. I removed the ' from "don't" and its working now, thanks!

Antoine Gaia Over a year ago

great, so could you accept the answer if it helps you ?

harshil9968 · Accepted Answer · 2016-12-14 09:21:12Z

0

from future import print_function
from os.path import join, dirname, abspath, isfile
from collections import Counter
import xlrd
from xlrd.sheet import ctype_text   


def get_excel_sheet_object(fname, idx=0):
    if not isfile(fname):
        print ('File doesn't exist: ', fname)
            # Open the workbook and 1st sheet
    xl_workbook = xlrd.open_workbook(fname)
    xl_sheet = xl_workbook.sheet_by_index(0)
    print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)

    return xl_sheet

xl_sheet_obj = get_excel_sheet_object('FILE_NAME_HERE')

Do anything you want with xl_sheet_obj after that this object is the excel sheet object.

answered Dec 14, 2016 at 9:21

harshil9968

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