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I wanted my code to have a button that lets me select the file I wanted to use instead of manually typing the file name in the code. My file is the file1234 as you can see below from my code. I also wanted to use ipywidgets if possible.

from openpyxl import Workbook

wb = Workbook()

with open('file1234.txt', encoding='utf-8') as blah:
    row = 1
    column = 1
    ws = wb.active
    
    lines = [line for line in blah.readlines()
            if 'entered the channel' not in line
            and 'left the channel' not in line]
    
    for line in lines:
        if column == 1:
            ## split the line and rejoin
            value = " ".join(line.strip().split(' ')[2:])
        else:
            value = line.strip()
            
        ws.cell(row=row, column=column, value=value)
        
        if (column := column + 1) > 3:
            row += 1
            column = 1
            
    for column_cells in ws.columns:
        length = max(len(str(cell.value)) for cell in column_cells)
        ws.column_dimensions[column_cells[0].column_letter].width = length

    wb.save('txt_toexl.xlsx')
  • 1
    Does this answer your question? [Quick and easy file dialog in Python?](https://stackoverflow.com/questions/9319317/quick-and-easy-file-dialog-in-python) – G. Anderson Mar 01 '22 at 23:34
  • Hi, I don't think that answers my question :'( –  Mar 01 '22 at 23:44

0 Answers0