How do I iterate over my sheet and color in cells that meet a specific condition? Using Python and openpyxl

Question

I want to iterate over a sheet and fill in cells if the cell is not 'empty'/null
if the cell is empty then no action required
if the cell is !empty, then color (fill) it in

the error message is

sheet[i].fill = PatternFill(fill_type='solid', start_color='43e33b', end_color='43e33b')
AttributeError: 'tuple' object has no attribute 'fill'

The code I'm using is

import openpyxl as xl

from openpyxl.styles import Font, Fill, Color, colors, PatternFill

# creating a variable to take any filename input from user
filename = input('Enter filename here: ')

# loading workbook on local computer c drive using filename
wb = xl.load_workbook(f'c:\\Users\\Charlie\\Desktop\\{filename}.xlsx')

# working with sheet1 on wb 'workbook'
sheet = wb['Sheet1']


# defining pacing list function
def packing_list():
    # deleting columns so that columns required are left for new file
    sheet.delete_cols(1, 14)
    max_rows = sheet.max_row
    print(max_rows)
    for i in range(1, max_rows):
        if sheet.cell(column=7, row=i, value=""):
            sheet[i].fill = PatternFill(fill_type='solid', 
            start_color='43e33b', end_color='43e33b')
        else:
            break
    # saving new worksheet to desktop with name packing_list
    wb.save('c:\\Users\\Charlie\\Desktop\\packing_list.xlsx')

Answer 1

sheet[i] is a tuple containing all the cells in the ith row, where i >= 1

sheet[i][j] is the i,jth cell, where j >= 0. So the 7th column would be sheet[i][6].

You could get the same thing using sheet.cell(column=7, row=i), where both row and column are 1-based.

You should be able to set sheet.cell(column=7, row=i).fill without any problems