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How to get file from url in python?

I want to download text files using python, how can I do so?
I used requests module's urlopen(url).read() but it gives me the bytes representation of file.

For me, I had to do the following (Python 3):
from urllib.request import urlopen
data = urlopen("[your url goes here]").read().decode('utf-8')
# Do what you need to do with the data.

You can use multiple options:
For the simpler solution you can use this
file_url = 'https://someurl.com/text_file.txt'
for line in urllib.request.urlopen(file_url):
print(line.decode('utf-8'))
For an API solution
file_url = 'https://someurl.com/text_file.txt'
response = requests.get(file_url)
if (response.status_code):
data = response.text
for line in enumerate(data.split('\n')):
print(line)

When downloading text files with python I like to use the wget module
import wget
remote_url = 'https://www.google.com/test.txt'
local_file = 'local_copy.txt'
wget.download(remote_url, local_file)
If that doesn't work try using urllib
from urllib import request
remote_url = 'https://www.google.com/test.txt'
file = 'copy.txt'
request.urlretrieve(remote_url, file)
When you are using the request module you are reading the file directly from the internet and it is causing you to see the text in byte format. Try to write the text to a file then view it manually by opening it on your desktop
import requests
remote_url = 'test.com/test.txt'
local_file = 'local_file.txt'
data = requests.get(remote_url)
with open(local_file, 'wb')as file:
file.write(data.content)

python uploading a remote file to GCS , without saving it in the machine [duplicate]

I have managed to get my first python script to work which downloads a list of .ZIP files from a URL and then proceeds to extract the ZIP files and writes them to disk.
I am now at a loss to achieve the next step.
My primary goal is to download and extract the zip file and pass the contents (CSV data) via a TCP stream. I would prefer not to actually write any of the zip or extracted files to disk if I could get away with it.
Here is my current script which works but unfortunately has to write the files to disk.
import urllib, urllister
import zipfile
import urllib2
import os
import time
import pickle
# check for extraction directories existence
if not os.path.isdir('downloaded'):
os.makedirs('downloaded')
if not os.path.isdir('extracted'):
os.makedirs('extracted')
# open logfile for downloaded data and save to local variable
if os.path.isfile('downloaded.pickle'):
downloadedLog = pickle.load(open('downloaded.pickle'))
else:
downloadedLog = {'key':'value'}
# remove entries older than 5 days (to maintain speed)
# path of zip files
zipFileURL = "http://www.thewebserver.com/that/contains/a/directory/of/zip/files"
# retrieve list of URLs from the webservers
usock = urllib.urlopen(zipFileURL)
parser = urllister.URLLister()
parser.feed(usock.read())
usock.close()
parser.close()
# only parse urls
for url in parser.urls:
if "PUBLIC_P5MIN" in url:
# download the file
downloadURL = zipFileURL + url
outputFilename = "downloaded/" + url
# check if file already exists on disk
if url in downloadedLog or os.path.isfile(outputFilename):
print "Skipping " + downloadURL
continue
print "Downloading ",downloadURL
response = urllib2.urlopen(downloadURL)
zippedData = response.read()
# save data to disk
print "Saving to ",outputFilename
output = open(outputFilename,'wb')
output.write(zippedData)
output.close()
# extract the data
zfobj = zipfile.ZipFile(outputFilename)
for name in zfobj.namelist():
uncompressed = zfobj.read(name)
# save uncompressed data to disk
outputFilename = "extracted/" + name
print "Saving extracted file to ",outputFilename
output = open(outputFilename,'wb')
output.write(uncompressed)
output.close()
# send data via tcp stream
# file successfully downloaded and extracted store into local log and filesystem log
downloadedLog[url] = time.time();
pickle.dump(downloadedLog, open('downloaded.pickle', "wb" ))

Below is a code snippet I used to fetch zipped csv file, please have a look:
Python 2:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(StringIO(resp.read()))
for line in myzip.open(file).readlines():
print line
Python 3:
from io import BytesIO
from zipfile import ZipFile
from urllib.request import urlopen
# or: requests.get(url).content
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(BytesIO(resp.read()))
for line in myzip.open(file).readlines():
print(line.decode('utf-8'))
Here file is a string. To get the actual string that you want to pass, you can use zipfile.namelist(). For instance,
resp = urlopen('http://mlg.ucd.ie/files/datasets/bbc.zip')
myzip = ZipFile(BytesIO(resp.read()))
myzip.namelist()
# ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']

My suggestion would be to use a StringIO object. They emulate files, but reside in memory. So you could do something like this:
# get_zip_data() gets a zip archive containing 'foo.txt', reading 'hey, foo'
import zipfile
from StringIO import StringIO
zipdata = StringIO()
zipdata.write(get_zip_data())
myzipfile = zipfile.ZipFile(zipdata)
foofile = myzipfile.open('foo.txt')
print foofile.read()
# output: "hey, foo"
Or more simply (apologies to Vishal):
myzipfile = zipfile.ZipFile(StringIO(get_zip_data()))
for name in myzipfile.namelist():
[ ... ]
In Python 3 use BytesIO instead of StringIO:
import zipfile
from io import BytesIO
filebytes = BytesIO(get_zip_data())
myzipfile = zipfile.ZipFile(filebytes)
for name in myzipfile.namelist():
[ ... ]

I'd like to offer an updated Python 3 version of Vishal's excellent answer, which was using Python 2, along with some explanation of the adaptations / changes, which may have been already mentioned.
from io import BytesIO
from zipfile import ZipFile
import urllib.request
url = urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/loc162txt.zip")
with ZipFile(BytesIO(url.read())) as my_zip_file:
for contained_file in my_zip_file.namelist():
# with open(("unzipped_and_read_" + contained_file + ".file"), "wb") as output:
for line in my_zip_file.open(contained_file).readlines():
print(line)
# output.write(line)
Necessary changes:
There's no StringIO module in Python 3 (it's been moved to io.StringIO). Instead, I use io.BytesIO]2, because we will be handling a bytestream -- Docs, also this thread.
urlopen:
"The legacy urllib.urlopen function from Python 2.6 and earlier has been discontinued; urllib.request.urlopen() corresponds to the old urllib2.urlopen.", Docs and this thread.
Note:
In Python 3, the printed output lines will look like so: b'some text'. This is expected, as they aren't strings - remember, we're reading a bytestream. Have a look at Dan04's excellent answer.
A few minor changes I made:
I use with ... as instead of zipfile = ... according to the Docs.
The script now uses .namelist() to cycle through all the files in the zip and print their contents.
I moved the creation of the ZipFile object into the with statement, although I'm not sure if that's better.
I added (and commented out) an option to write the bytestream to file (per file in the zip), in response to NumenorForLife's comment; it adds "unzipped_and_read_" to the beginning of the filename and a ".file" extension (I prefer not to use ".txt" for files with bytestrings). The indenting of the code will, of course, need to be adjusted if you want to use it.
Need to be careful here -- because we have a byte string, we use binary mode, so "wb"; I have a feeling that writing binary opens a can of worms anyway...
I am using an example file, the UN/LOCODE text archive:
What I didn't do:
NumenorForLife asked about saving the zip to disk. I'm not sure what he meant by it -- downloading the zip file? That's a different task; see Oleh Prypin's excellent answer.
Here's a way:
import urllib.request
import shutil
with urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/2015-2_UNLOCODE_SecretariatNotes.pdf") as response, open("downloaded_file.pdf", 'w') as out_file:
shutil.copyfileobj(response, out_file)

I'd like to add my Python3 answer for completeness:
from io import BytesIO
from zipfile import ZipFile
import requests
def get_zip(file_url):
url = requests.get(file_url)
zipfile = ZipFile(BytesIO(url.content))
files = [zipfile.open(file_name) for file_name in zipfile.namelist()]
return files.pop() if len(files) == 1 else files

write to a temporary file which resides in RAM
it turns out the tempfile module ( http://docs.python.org/library/tempfile.html ) has just the thing:
tempfile.SpooledTemporaryFile([max_size=0[,
mode='w+b'[, bufsize=-1[, suffix=''[,
prefix='tmp'[, dir=None]]]]]])
This
function operates exactly as
TemporaryFile() does, except that data
is spooled in memory until the file
size exceeds max_size, or until the
file’s fileno() method is called, at
which point the contents are written
to disk and operation proceeds as with
TemporaryFile().
The resulting file has one additional
method, rollover(), which causes the
file to roll over to an on-disk file
regardless of its size.
The returned object is a file-like
object whose _file attribute is either
a StringIO object or a true file
object, depending on whether
rollover() has been called. This
file-like object can be used in a with
statement, just like a normal file.
New in version 2.6.
or if you're lazy and you have a tmpfs-mounted /tmp on Linux, you can just make a file there, but you have to delete it yourself and deal with naming

Adding on to the other answers using requests:
# download from web
import requests
url = 'http://mlg.ucd.ie/files/datasets/bbc.zip'
content = requests.get(url)
# unzip the content
from io import BytesIO
from zipfile import ZipFile
f = ZipFile(BytesIO(content.content))
print(f.namelist())
# outputs ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
Use help(f) to get more functions details for e.g. extractall() which extracts the contents in zip file which later can be used with with open.

All of these answers appear too bulky and long. Use requests to shorten the code, e.g.:
import requests, zipfile, io
r = requests.get(zip_file_url)
z = zipfile.ZipFile(io.BytesIO(r.content))
z.extractall("/path/to/directory")

Vishal's example, however great, confuses when it comes to the file name, and I do not see the merit of redefing 'zipfile'.
Here is my example that downloads a zip that contains some files, one of which is a csv file that I subsequently read into a pandas DataFrame:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
import pandas
url = urlopen("https://www.federalreserve.gov/apps/mdrm/pdf/MDRM.zip")
zf = ZipFile(StringIO(url.read()))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])
(Note, I use Python 2.7.13)
This is the exact solution that worked for me. I just tweaked it a little bit for Python 3 version by removing StringIO and adding IO library
Python 3 Version
from io import BytesIO
from zipfile import ZipFile
import pandas
import requests
url = "https://www.nseindia.com/content/indices/mcwb_jun19.zip"
content = requests.get(url)
zf = ZipFile(BytesIO(content.content))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])

It wasn't obvious in Vishal's answer what the file name was supposed to be in cases where there is no file on disk. I've modified his answer to work without modification for most needs.
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
def unzip_string(zipped_string):
unzipped_string = ''
zipfile = ZipFile(StringIO(zipped_string))
for name in zipfile.namelist():
unzipped_string += zipfile.open(name).read()
return unzipped_string

Use the zipfile module. To extract a file from a URL, you'll need to wrap the result of a urlopen call in a BytesIO object. This is because the result of a web request returned by urlopen doesn't support seeking:
from urllib.request import urlopen
from io import BytesIO
from zipfile import ZipFile
zip_url = 'http://example.com/my_file.zip'
with urlopen(zip_url) as f:
with BytesIO(f.read()) as b, ZipFile(b) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read())
If you already have the file downloaded locally, you don't need BytesIO, just open it in binary mode and pass to ZipFile directly:
from zipfile import ZipFile
zip_filename = 'my_file.zip'
with open(zip_filename, 'rb') as f:
with ZipFile(f) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read().decode('utf-8'))
Again, note that you have to open the file in binary ('rb') mode, not as text or you'll get a zipfile.BadZipFile: File is not a zip file error.
It's good practice to use all these things as context managers with the with statement, so that they'll be closed properly.

How to translate url encoded string python

This code is supposed to download a list of pdfs into a directory
for pdf in preTag:
pdfUrl = "https://the-eye.eu/public/Books/Programming/" +
pdf.get("href")
print("Downloading...%s"% pdfUrl)
#downloading pdf from url
page = requests.get(pdfUrl)
page.raise_for_status()
#saving pdf to new directory
pdfFile = open(os.path.join(filePath, os.path.basename(pdfUrl)), "wb")
for chunk in page.iter_content(1000000):
pdfFile.write(chunk)
pdfFile.close()
I used os.path.basename() just to make sure the files would actually download. However, I want to know how to change the file name from 3D%20Printing%20Blueprints%20%5BeBook%5D.pdf to something like "3D Printing Blueprints.pdf"

You can use the urllib2 unquote function:
import urllib2
print urllib2.unquote("3D%20Printing%20Blueprints%20%5BeBook%5D.pdf") #3D Printing Blueprints.pdf

use this:
os.rename("3D%20Printing%20Blueprints%20%5BeBook%5D.pdf", "3D Printing Blueprints.pdf")
you can find more info here

How to download FTP file using its full FTP path?

Using the ftplib in Python, you can download files, but it seems you are restricted to use the file name only (not the full file path). The following code successfully downloads the requested code:
import ftplib
ftp=ftplib.FTP("ladsweb.nascom.nasa.gov")
ftp.login()
ftp.cwd("/allData/5/MOD11A1/2002/001")
ftp.retrbinary('RETR MOD11A1.A2002001.h00v08.005.2007079015634.hdf',open("MOD11A1.A2002001.h00v08.005.2007079015634.hdf",'wb').write)
As you can see, first a login to the site (ftp.login()) is established and then the current directory is set (ftp.cwd()). After that you need to declare the file name to download the file that resides in the current directory.
How about downloading the file directly by using its full path/link?

import ftplib
ftp = ftplib.FTP("ladsweb.nascom.nasa.gov")
ftp.login()
a = 'allData/5/MOD11A1/2002/001/MOD11A1.A2002001.h00v08.005.2007079015634.hdf'
fhandle = open('ftp-test', 'wb')
ftp.retrbinary('RETR ' + a, fhandle.write)
fhandle.close()

This solution uses the urlopen function in the urllib module. The urlopen function will let you download ftp and http urls. I like using it because you can connect and get all the data in one line. The last three lines extract the filename from the url and then save the data to that filename.
from urllib import urlopen
url = 'ftp://ladsweb.nascom.nasa.gov/allData/5/MOD11A1/2002/001/MOD11A1.A2002001.h00v08.005.2007079015634.hdf'
data = urlopen(url).read()
filename = url.split('/')[-1]
with open(filename, 'wb') as f:
f.write(data)

Saving a downloaded ZIP file w/Python

I'm working on a script that will automatically update an installed version of Calibre. Currently I have it downloading the latest portable version. I seem to be having trouble saving the zipfile. Currently my code is:
import urllib2
import re
import zipfile
#tell the user what is happening
print("Calibre is Updating")
#download the page
url = urllib2.urlopen ( "http://sourceforge.net/projects/calibre/files" ).read()
#determin current version
result = re.search('title="/[0-9.]*/([a-zA-Z\-]*-[0-9\.]*)', url).groups()[0][:-1]
#download file
download = "http://status.calibre-ebook.com/dist/portable/" + result
urllib2.urlopen( download )
#save
output = open('install.zip', 'w')
output.write(zipfile.ZipFile("install.zip", ""))
output.close()

You don't need to use zipfile.ZipFile for this (and the way you're using it, as well as urllib2.urlopen, has problems as well). Instead, you need to save the urlopen result in a variable, then read it and write that output to a .zip file. Try this code:
#download file
download = "http://status.calibre-ebook.com/dist/portable/" + result
request = urllib2.urlopen( download )
#save
output = open("install.zip", "w")
output.write(request.read())
output.close()

There also can be a one-liner:
open('install.zip', 'wb').write(urllib.urlopen('http://status.calibre-ebook.com/dist/portable/' + result).read())
which doesn't have a good memory-efficiency, but still works.

If you just want to download a file from the net, you can use urllib.urlretrieve:
Copy a network object denoted by a URL to a local file ...
Example using requests instead of urllib2:
import requests, re, urllib
print("Calibre is updating...")
content = requests.get("http://sourceforge.net/projects/calibre/files").content
# determine current version
v = re.search('title="/[0-9.]*/([a-zA-Z\-]*-[0-9\.]*)', content).groups()[0][:-1]
download_url = "http://status.calibre-ebook.com/dist/portable/{0}".format(v)
print("Downloading {0}".format(download_url))
urllib.urlretrieve(download_url, 'install.zip')
# file should be downloaded at this point

have you tryed
output = open('install.zip', 'wb') // note the "b" flag which means "binary file"