使用 Python 打开 Excel 工作簿

Question

I am learning Python through 'Automate the Boring Stuff With Python' First Edition. In chapter 12, pg 267, we are supposed to open a file called example.xlsx.

The author's code reads:

import openpyxl
wb = openpyxl.load_workbook('example.xlsx')
type(wb)

However, when I try to open this file, I get the following error (this is the last line of the error):

 FileNotFoundError: [Errno 2] No such file or directory: 'example.xlsx'

I know this file exists, because I downloaded it myself and am looking at it right now.

I have tried moving it to the current location in which Python 3.8 is, I have tried saving it with my Automate the Boring Stuff files that I've been working on the desktop, and I have tried saving it in every conceivable location on my machine, but I continue getting this same message.

I have imported openpyxl without error, but when I enter the line

wb = openpyxl.load_workbook('example.xlsx') 

I have entered the entire pathway for the example.xlsx in the parenthesis, and I continue to get the same error.

What am I doing wrong? How am I supposed to open an Excel workbook?

I still don't understand how I am doing wrong, but this one is incredibly infuriating, and I feel incredibly stupid, because it must be something simple.

Any insight/help is greatly appreciated.

Answer 1

Your error is unambigous — your file in a supposed directory don't exist . Believe me.

For Python is irrelevant, whether you see it. Python itself must see it.

Specify the full path, using forward slashes, for example:

wb = openpyxl.load_workbook('C:/users/John/example.xlsx')

---

Or find out your real current (working) directory — and not the one supposed by you — with commands

import os
print(os.getcwd())

then move your example.xlsx to it, and then use only the name of your file

wb = openpyxl.load_workbook('example.xlsx')

---

You may also verify its existence with commands — use copy/paste from your code to avoid typos in the file name / path

import os.path
print(os.path.exists('example.xlsx'))                   # True, if Python sees it

or

import os.path
print(os.path.exists('C:/users/John/example.xlsx'))     # True, if exists

to be sure that I'm right, ie that the error is not in the function openpyxl.load_workbook() itself, but in its parameter (the path to the file) provided by you .

Answer 2

I notice that the extension of the example file is not the same as described in the book, is example.csv. I was facing the same frustration as you