Given a URL to a text file, what is the simplest way to read the contents of the text file?

Question

In Python, when given the URL for a text file, what is the simplest way to access the contents off the text file and print the contents of the file out locally line-by-line without saving a local copy of the text file?

TargetURL=http://www.myhost.com/SomeFile.txt
#read the file
#print first line
#print second line
#etc

Asked By: Chris

||

Source

Answer 1

import urllib2

f = urllib2.urlopen(target_url)
for l in f.readlines():
    print l

Answered By: Will

Answer 2

import urllib2
for line in urllib2.urlopen("http://www.myhost.com/SomeFile.txt"):
    print line

Answered By: Fabian

Answer 3

Edit 09/2016: In Python 3 and up use urllib.request instead of urllib2

Actually the simplest way is:

import urllib2  # the lib that handles the url stuff

data = urllib2.urlopen(target_url) # it's a file like object and works just like a file
for line in data: # files are iterable
    print line

You don’t even need "readlines", as Will suggested. You could even shorten it to: ^*

import urllib2

for line in urllib2.urlopen(target_url):
    print line

But remember in Python, readability matters.

However, this is the simplest way but not the safe way because most of the time with network programming, you don’t know if the amount of data to expect will be respected. So you’d generally better read a fixed and reasonable amount of data, something you know to be enough for the data you expect but will prevent your script from been flooded:

import urllib2

data = urllib2.urlopen("http://www.google.com").read(20000) # read only 20 000 chars
data = data.split("n") # then split it into lines

for line in data:
    print line

^{* Second example in Python 3:}

import urllib.request  # the lib that handles the url stuff

for line in urllib.request.urlopen(target_url):
    print(line.decode('utf-8')) #utf-8 or iso8859-1 or whatever the page encoding scheme is

Answered By: e-satis

Answer 4

There’s really no need to read line-by-line. You can get the whole thing like this:

import urllib
txt = urllib.urlopen(target_url).read()

Answered By: Ken Kinder

Answer 5

I’m a newbie to Python and the offhand comment about Python 3 in the accepted solution was confusing. For posterity, the code to do this in Python 3 is

import urllib.request
data = urllib.request.urlopen(target_url)

for line in data:
    ...

or alternatively

from urllib.request import urlopen
data = urlopen(target_url)

Note that just import urllib does not work.

Answered By: Andrew Mao

Answer 6

Another way in Python 3 is to use the urllib3 package.

import urllib3

http = urllib3.PoolManager()
response = http.request('GET', target_url)
data = response.data.decode('utf-8')

This can be a better option than urllib since urllib3 boasts having

Thread safety.

Connection pooling.

Client-side SSL/TLS verification.

File uploads with multipart encoding.

Helpers for retrying requests and dealing with HTTP redirects.

Support for gzip and deflate encoding.

Proxy support for HTTP and SOCKS.

100% test coverage.

Answered By: leafmeal

Answer 7

The requests library has a simpler interface and works with both Python 2 and 3.

import requests

response = requests.get(target_url)
data = response.text

Answered By: leafmeal

Answer 8

For me, none of the above responses worked straight ahead. Instead, 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.

Answered By: bmiselis

Answer 9

Just updating here the solution suggested by @ken-kinder for Python 2 to work with Python 3:

import urllib
urllib.request.urlopen(target_url).read()

Answered By: delimiter

Answer 10

requests package works really well for simple ui
as @Andrew Mao suggested

import requests
response = requests.get('http://lib.stat.cmu.edu/datasets/boston')
data = response.text
for i, line in enumerate(data.split('n')):
    print(f'{i}   {line}')

o/p:

0    The Boston house-price data of Harrison, D. and Rubinfeld, D.L. 'Hedonic
1    prices and the demand for clean air', J. Environ. Economics & Management,
2    vol.5, 81-102, 1978.   Used in Belsley, Kuh & Welsch, 'Regression diagnostics
3    ...', Wiley, 1980.   N.B. Various transformations are used in the table on
4    pages 244-261 of the latter.
5   
6    Variables in order:

Checkout kaggle notebook on how to extract dataset/dataframe from URL

Answered By: arun

Answer 11

I do think requests is the best option. Also note the possibility of setting encoding manually.

import requests
response = requests.get("http://www.gutenberg.org/files/10/10-0.txt")
# response.encoding = "utf-8"
hehe = response.text

Answered By: xiaoou wang

Answer 12

You can use this, as well for simple methodology:

import requests
url_res = requests.get(url= "http://www.myhost.com/SomeFile.txt")
with open(filename + ".txt", "wb") as file:
    file.write(url_res.content)

Answered By: Abhishek Raj

Answer 13

None of these answers work in Python 3. I am using Python 3.9 and refuse to import urllib2 which dates back to at least 2010.

Here is how you read a text file located on a remote server given the url:

import io
import urllib
hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
   'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
   'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
   'Accept-Encoding': 'none',
   'Accept-Language': 'en-US,en;q=0.8',
   'Connection': 'keep-alive'}
url = 'https://server.com/path/hello world.txt'

req = urllib.request.Request(url, headers=hdr)
u = urllib.request.urlopen(req)
file = io.TextIOWrapper(u, encoding='utf-8')
file_contents = file.read()
print(file_contents)

Hope this is helpful to someone because it was very hard to find the answer.

Answered By: Jim Dandy BOA

Given a URL to a text file, what is the simplest way to read the contents of the text file?

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