Download csv file through python (url)

Question

I work on a project and I want to download a csv file from a url. I did some research on the site but none of the solutions presented worked for me.

The url offers you directly to download or open the file of the blow I do not know how to say a python to save the file (it would be nice if I could also rename it)

But when I open the url with this code nothing happens.

import urllib
url='https://data.toulouse-metropole.fr/api/records/1.0/download/?dataset=dechets-menagers-et-assimiles-collectes'

testfile = urllib.request.urlopen(url)

Any ideas?

Have you taken a look at this question: http://stackoverflow.com/questions/19602931/basic-http-file-downloading-and-saving-to-disk-in-python — roach, Jan 02 '17 at 23:16
Yes! the answers is : testfile = urllib.URLopener() testfile.retrieve("http://randomsite.com/file.gz", "file.gz") but i got errors: 'HTTPResponse' object has no attribute 'retrieve' — blabla, Jan 02 '17 at 23:21

score 14 · Accepted Answer · answered Jan 02 '17 at 23:22

Try this. Change "folder" to a folder on your machine

import os
import requests

url='https://data.toulouse-metropole.fr/api/records/1.0/download/?dataset=dechets-menagers-et-assimiles-collectes'
response = requests.get(url)
with open(os.path.join("folder", "file"), 'wb') as f:
    f.write(response.content)

score 1 · Answer 2 · answered Jan 02 '17 at 23:30

You can adapt an example from the docs

import urllib.request
url='https://data.toulouse-metropole.fr/api/records/1.0/download/?dataset=dechets-menagers-et-assimiles-collectes'

with urllib.request.urlopen(url) as testfile, open('dataset.csv', 'w') as f:
    f.write(testfile.read().decode())

Download csv file through python (url)

2 Answers2