'To unpack a zip file from url request

I want to get this XML file from a danish website with the URL https://www.foedevarestyrelsen.dk/_layouts/15/sdata/smileystatus.zip, but the data comes in a zip file. This is problematic because I want to get the data sheet directly into Python for a school project, and not to save it local and convert it into csv. So far I have tried following codes inspired from answers to previous questions about this subject....

from zipfile import ZipFile
from io import BytesIO
import requests
import pandas as pd

def get_zip(file_url):
    url = requests.get(file_url)
    zipfile = ZipFile(BytesIO(url.content))
    zip_names = zipfile.namelist()
    if len(zip_names) == 1:
        SmileyStatus = zip_names.pop()
        extracted_file = zipfile.open("SmileyStatus.xls")
        return extracted_file

xls = get_zip("https://www.foedevarestyrelsen.dk/_layouts/15/sdata/smileystatus.zip")

data = pd.read_excel(xls)

and I have also tried:

from zipfile import ZipFile
import requests
import pandas as pd

url = 'https://www.foedevarestyrelsen.dk/_layouts/15/sdata/smileystatus.zip'
df = pd.read_csv(url, 
                 compression = "zip")

Solution 1:^[1]

Know its an old post, but I just had a similar need and found a good solution.

import pandas as pd

import io

import requests

url = 'https://whatever.website/some_file.zip'

r = requests.get(url)

df = pd.read_csv(io.BytesIO(r.content), compression="zip")

print(df.head())

print(f"{len(df)} rows")

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Source: Stack Overflow

Solution	Source
Solution 1	Craig9d

'To unpack a zip file from url request

Solution 1:[1]

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Solution 1:^[1]