Is it possible to load xlsx with openpyxl from zipfile - excel

Question

I'm trying to openpyxl.load_workbook xlsx files from compressed zip file, but it doesn't work. The following code fails at openpyxl.load_workbook with "BadZipfile: File is not a zip file"
with zipfile.ZipFile(os.path.join(root, raw)) as z:
for file_info in z.infolist():
wb = openpyxl.load_workbook(z.open(file_info), read_only=True)
There is nothing wrong with the archive and the excel file in it, as if i extract it to disk then the following works:
with open('report.xlsx') as f:
wb = openpyxl.load_workbook(f, read_only=True)
I can go with this solution and temporary extract it somewhere and load xslx, but would like to understand if it possible to load it from zipfile.

Answer 1

The problem is that readonly=True does not do quite what you think it does. According to the docs:
Fortunately, there are two modes that enable you to read and write unlimited amounts of data with (near) constant memory consumption.
While not explicitly stated, I would assume that this involves some equivalent to a memory-mapped file (because of "constant memory consumption") and random access (because of the range of allowed operations).
Either way, setting readonly=True is not an indication of the fact that you only intend to read a workbook (that's all load_workbook can do anyway, you have to overwrite the existing one to make any "changes"). It is an indication of the fact that you want to access the file directly on disk, without loading the entire contents.
It seems pretty clear (and intuitively expected) that ZipFile.open does not provide a random-access file:
Note: The file-like object is read-only and provides the following methods: read(), readline(), readlines(), __iter__(), next().
The fact that seek is not mentioned in this list is quite telling (pun only somewhat intended).
You can get more information about the exception by splitting the offending line into two (a useful general debugging technique for nested function calls):
x = z.open(file_info)
wb = openpyxl.load_workbook(x, readonly=True)
You will notice that the error occurs on the second of those two lines. This is because pretty much all the Microsoft open-document formats are actually just fancy zip files. The problem is most likely that openpyxl cannot open your file in random access mode, not that it's actually an invalid zip file.
Either way, this is a bunch of very educated guesswork that leads to a simple, one-keyword-deletion solution:
TL;DR
Get rid of readonly=True when reading non-random-access data like a compressed zip entry:
wb = openpyxl.load_workbook(z.open(file_info))
Appendix
You should get in the habit of writing minimal programs that demonstrate your issue so that people answering your question can focus on doing their job instead of getting irritated and closing down what would otherwise be a perfectly good question. I liked your question enough to do that for you, so here is a minimal program that demonstrates your issue and requires nothing more than copy-and-paste to run:
import openpyxl, zipfile
from openpyxl.workbook.workbook import Workbook
wb = Workbook()
wb.active['A1'] = 12
wb.active['A2'] = 13
wb.save('report.xlsx')
with zipfile.ZipFile('test.zip', 'w') as z:
z.write('report.xlsx')
with open('report.xlsx') as f:
wb = openpyxl.load_workbook(f, read_only=True)
print(wb.active['A1'].value)
print(wb.active['A2'].value)
with zipfile.ZipFile('test.zip', 'r') as z:
for file_info in z.infolist():
x = z.open(file_info, 'r')
wb = openpyxl.load_workbook(x, readonly=True)
print(wb.active['A1'].value)
print(wb.active['A2'].value)