What is the quickest way to HTTP GET in Python?

Question

What is the quickest way to HTTP GET in Python if I know the content will be a string? I am searching the documentation for a quick one-liner like:

contents = url.get("http://example.com/foo/bar")

But all I can find using Google are httplib and urllib - and I am unable to find a shortcut in those libraries.

Does standard Python 2.5 have a shortcut in some form as above, or should I write a function url_get?

1. I would prefer not to capture the output of shelling out to wget or curl.

Answer 1

Python 3:

import urllib.request
contents = urllib.request.urlopen("http://example.com/foo/bar").read()

Python 2:

import urllib2
contents = urllib2.urlopen("http://example.com/foo/bar").read()

Documentation for urllib.request and read.

Answer 2

You could use a library called requests.

import requests
r = requests.get("http://example.com/foo/bar")

This is quite easy. Then you can do like this:

>>> print(r.status_code)
>>> print(r.headers)
>>> print(r.content)

Answer 3

If you want solution with httplib2 to be oneliner consider instantiating anonymous Http object

import httplib2
resp, content = httplib2.Http().request("http://example.com/foo/bar")

Answer 4

Have a look at httplib2, which - next to a lot of very useful features - provides exactly what you want.

import httplib2

resp, content = httplib2.Http().request("http://example.com/foo/bar")

Where content would be the response body (as a string), and resp would contain the status and response headers.

It doesn't come included with a standard python install though (but it only requires standard python), but it's definitely worth checking out.

Answer 5

It's simple enough with the powerful urllib3 library.

Import it like this:

import urllib3

http = urllib3.PoolManager()

And make a request like this:

response = http.request('GET', 'https://example.com')

print(response.data) # Raw data.
print(response.data.decode('utf-8')) # Text.
print(response.status) # Status code.
print(response.headers['Content-Type']) # Content type.

You can add headers too:

response = http.request('GET', 'https://example.com', headers={
    'key1': 'value1',
    'key2': 'value2'
})

More info can be found on the urllib3 documentation.

urllib3 is much safer and easier to use than the builtin urllib.request or http modules and is stable.

Answer 6

Without further necessary imports this solution works (for me) - also with https:

try:
    import urllib2 as urlreq # Python 2.x
except:
    import urllib.request as urlreq # Python 3.x
req = urlreq.Request("http://example.com/foo/bar")
req.add_header('User-Agent', 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/60.0.3112.113 Safari/537.36')
urlreq.urlopen(req).read()

I often have difficulty grabbing the content when not specifying a "User-Agent" in the header information. Then usually the requests are cancelled with something like: urllib2.HTTPError: HTTP Error 403: Forbidden or urllib.error.HTTPError: HTTP Error 403: Forbidden.

Answer 7

theller's solution for wget is really useful, however, i found it does not print out the progress throughout the downloading process. It's perfect if you add one line after the print statement in reporthook.

import sys, urllib

def reporthook(a, b, c):
    print "% 3.1f%% of %d bytes\r" % (min(100, float(a * b) / c * 100), c),
    sys.stdout.flush()
for url in sys.argv[1:]:
    i = url.rfind("/")
    file = url[i+1:]
    print url, "->", file
    urllib.urlretrieve(url, file, reporthook)
print

Answer 8

Here is a wget script in Python:

# From python cookbook, 2nd edition, page 487
import sys, urllib

def reporthook(a, b, c):
    print "% 3.1f%% of %d bytes\r" % (min(100, float(a * b) / c * 100), c),
for url in sys.argv[1:]:
    i = url.rfind("/")
    file = url[i+1:]
    print url, "->", file
    urllib.urlretrieve(url, file, reporthook)
print

Answer 9

How to also send headers

Python 3:

import urllib.request
contents = urllib.request.urlopen(urllib.request.Request(
    "https://api.github.com/repos/cirosantilli/linux-kernel-module-cheat/releases/latest",
    headers={"Accept" : 'application/vnd.github.full+json"text/html'}
)).read()
print(contents)

Python 2:

import urllib2
contents = urllib2.urlopen(urllib2.Request(
    "https://api.github.com",
    headers={"Accept" : 'application/vnd.github.full+json"text/html'}
)).read()
print(contents)

Answer 10

Actually in Python we can read from HTTP responses like from files, here is an example for reading JSON from an API.

import json
from urllib.request import urlopen

with urlopen(url) as f:
    resp = json.load(f)

return resp['some_key']

Answer 11

If you are working with HTTP APIs specifically, there are also more convenient choices such as Nap.

For example, here's how to get gists from Github since May 1st 2014:

from nap.url import Url
api = Url('https://api.github.com')

gists = api.join('gists')
response = gists.get(params={'since': '2014-05-01T00:00:00Z'})
print(response.json())

More examples: https://github.com/kimmobrunfeldt/nap#examples

Answer 12

Excellent solutions Xuan, Theller.

For it to work with python 3 make the following changes

import sys, urllib.request

def reporthook(a, b, c):
    print ("% 3.1f%% of %d bytes\r" % (min(100, float(a * b) / c * 100), c))
    sys.stdout.flush()
for url in sys.argv[1:]:
    i = url.rfind("/")
    file = url[i+1:]
    print (url, "->", file)
    urllib.request.urlretrieve(url, file, reporthook)
print

Also, the URL you enter should be preceded by a "http://", otherwise it returns a unknown url type error.

Answer 13

For python >= 3.6, you can use dload:

import dload
t = dload.text(url)

---

For json:

j = dload.json(url)

---

Install:
pip install dload

Answer 14

If you want a lower level API:

import http.client

conn = http.client.HTTPSConnection('example.com')
conn.request('GET', '/')

resp = conn.getresponse()
content = resp.read()

conn.close()

text = content.decode('utf-8')

print(text)