I want to download zip file of this link. I tried various method but I couldn't do this.
url = "https://www.cms.gov/apps/ama/license.asp?file=http://download.cms.gov/Research-Statistics-Data-and-Systems/Statistics-Trends-and-Reports/Medicare-Provider-Charge-Data/Downloads/Medicare_National_HCPCS_Aggregate_CY2017.zip"
# downloading with requests
# import the requests library
import requests
# download the file contents in binary format
r = requests.get(url)
# open method to open a file on your system and write the contents
with open("minemaster1.zip", "wb") as code:
code.write(r.content)
# downloading with urllib
# import the urllib library
import urllib
# Copy a network object to a local file
urllib.request.urlretrieve(url, "minemaster.zip")
Can anybody help me in resolving this issue.
They're using some accept/decline mechanism, so you'll need to add this parameters to url:
url = 'http://download.cms.gov/Research-Statistics-Data-and-Systems/Statistics-Trends-and-Reports/Medicare-Provider-Charge-Data/Downloads/Medicare_National_HCPCS_Aggregate_CY2017.zip?agree=yes&next=Accept'
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I'm working on a program that downloads data from a series of URLs, like this:
https://server/api/getsensordetails.xmlid=sesnsorID&username=user&password=password
the program goes through a list with IDs (about 2500) and running the URL, try to do it using the following code
import webbrowser
webbrowser.open(url)
but this code implies to open the URL in the browser and confirm if I want to download, I need him to simply download the files without opening a browser and much less without having to confirm
thanks for everything
You can use the Requests library.
import requests
print('Beginning file download with requests')
url = 'http://PathToFile.jpg'
r = requests.get(url)
with open('pathOfFileToReceiveDownload.jpg', 'wb') as f:
f.write(r.content)
I have a URL which upon loading in, will automatically download a CSV to your machine.
I am trying to do this automatically in python and control what the downloaded file is named. Here is my current code:
import os
import sys
import urllib.request
URL= https://www.google.com/url?q=https%3A%2F%2Fbasketballmonster.com%2FDaily.aspx%3Fv%3D2%26exportcsv%3DXnZZUZaDa0E296JhVEGWbs8HRGOXsEkeJKs2towTT%2Fw%3D&sa=D&sntz=1&usg=AFQjCNHYm9T_QIZvEJ8qIKfyXQuZb4HPVA
response = urllib.request.urlopen(URL)
URL2 = response.geturl()
urllib.request.urlretrieve(URL2, "file2.csv")
For the URL:
https://www.google.com/url?q=https%3A%2F%2Fbasketballmonster.com%2FDaily.aspx%3Fv%3D2%26exportcsv%3DXnZZUZaDa0E296JhVEGWbs8HRGOXsEkeJKs2towTT%2Fw%3D&sa=D&sntz=1&usg=AFQjCNHYm9T_QIZvEJ8qIKfyXQuZb4HPVA
(clicking that downloads a CSV to disk)
However, the CSV downloaded has this html markup instead of being the actual data
Any ideas on a solution?
I'm writing a program/script in python3. I know how to download single files from URL, but I need to download whole folder, unzip the files and merge text files.
Is it possible to download all files FROM HERE to new folder on my computer with python? I'm using a urllib to download a single files, can anyone give a example how to download whole folder from link above?
Install bs4 and requests, than you can use code like this:
import bs4
import requests
url = "http://bossa.pl/pub/metastock/ofe/sesjaofe/"
r = requests.get(url)
data = bs4.BeautifulSoup(r.text, "html.parser")
for l in data.find_all("a"):
r = requests.get(url + l["href"])
print(r.status_code)
Than you have to save the data of the request into your directory.
I need to download a file from an external source, I am using Basic authentication to login to the URL
import requests
response = requests.get('<external url', auth=('<username>', '<password>'))
data = response.json()
html = data['list'][0]['attachments'][0]['url']
print (html)
data = requests.get('<API URL to download the attachment>', auth=('<username>', '<password>'), stream=True)
print (data.content)
I am getting below output
<url to download the binary data>
\x00\x00\x13\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x0f\xcb\x00\x00\x1e\x00\x1e\x00\xbe\x07\x00\x00.\xcf\x05\x00\x00\x00'
I am expecting the URL to download the word document within the same session.
Working solution
import requests
import shutil
response = requests.get('<url>', auth=('<username>', '<password>'))
data = response.json()
html = data['list'][0]['attachments'][0]['url']
print (html)
data = requests.get('<url>', auth=('<username>', '<password>'), stream=True)
with open("C:/myfile.docx", 'wb') as f:
data.raw.decode_content = True
shutil.copyfileobj(data.raw, f)
I am able to download the file as it is.
When you want to download a file directly you can use shutil.copyfileobj():
https://docs.python.org/2/library/shutil.html#shutil.copyfileobj
You already are passing stream=True to requests which is what you need to get a file-like object back. Just pass that as the source to copyfileobj().
I am trying to download a pdf from a webpage using urllib. I used the source link that downloads the file in the browser but that same link fails to download the file in Python. Instead what downloads is a redirect to the main page.
import os
import urllib
os.chdir(r'/Users/file')
url = "http://www.australianturfclub.com.au/races/SectionalsMeeting.aspx?meetingId=2414"
urllib.urlretrieve (url, "downloaded_file")
Please try downloading the file manually from the link provided or from the redirected site, the link on the main page is called 'sectionals'.
Your help is much appreciated.
It is because the given link redirects you to a "raw" pdf file. Examining the response headers via Firebug, I am able to get the filename sectionals/2014/2607RAND.pdf (see screenshot below) and as it is relative to the current .aspx file, the required URI should be switched to (in your case by changing the url variable to this link) http://www.australianturfclub.com.au/races/sectionals/2014/2607RAND.pdf
In python3:
import urllib.request
import shutil
local_filename, headers = urllib.request.urlretrieve('http://www.australianturfclub.com.au/races/SectionalsMeeting.aspx?meetingId=2414')
shutil.move(local_filename, 'ret.pdf')
The shutil is there because python save to a temp folder (im my case, that's another partition so os.rename will give me an error).