the url link is the direct link to a web file (xlsb file) which I am trying to downlead. The code below works with no error and the file seems created in the path but once I try to open it, corrupt file message pops up on excel. The response status is 400 so it is a bad request. Any advice on this?

url = 'http://rigcount.bakerhughes.com/static-files/55ff50da-ac65-410d-924c-fe45b23db298'
file_name = r'local path with xlsb extension'


with open(file_name, "wb") as file:
    response = requests.request(method="GET", url=url)
    file.write(response.content) 

CodePudding user response：

Seems working for me. Try this out:

from requests import get

url = 'http://rigcount.bakerhughes.com/static-files/55ff50da-ac65-410d-924c-fe45b23db298'

# make HTTP request to fetch data
r = get(url)

# check if request is success
r.raise_for_status()

# write out byte content to file
with open('out.xlsb', 'wb') as out_file:
    out_file.write(r.content)