CODEWITHSUNDEEP
pythonhttpnetwork-programming
Frank Krueger
Frank Krueger

Reputation: 71053

What is the quickest way to HTTP GET in Python?

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.

Upvotes: 766

Views: 1034492

Answers (14)

user1001237
user1001237

Reputation:

Use the Requests library:

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

Then you can do stuff like this:

>>> print(r.status_code)
>>> print(r.headers)
>>> print(r.content)  # bytes
>>> print(r.text)     # r.content as str

Install Requests by running this command:

pip install requests

Upvotes: 529

Katrych Taras
Katrych Taras

Reputation: 161

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']

Upvotes: 16

user10167940
user10167940

Reputation:

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)

Upvotes: 5

Pedro Lobito
Pedro Lobito

Reputation: 99061

For python >= 3.6, you can use dload:

import dload
t = dload.text(url)

For json:

j = dload.json(url)

Install:
pip install dload

Upvotes: 1

user10167940
user10167940

Reputation:

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.

Upvotes: 19

Nick Presta
Nick Presta

Reputation: 28705

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.

Upvotes: 1026

Ciro Santilli OurBigBook.com
Ciro Santilli OurBigBook.com

Reputation: 384394

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)

Upvotes: 7

to-chomik
to-chomik

Reputation: 371

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")

Upvotes: 31

michael_s
michael_s

Reputation: 2575

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.

Upvotes: 9

Steven
Steven

Reputation:

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.

Upvotes: 20

Akshar
Akshar

Reputation: 957

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.

Upvotes: 3

Kimmo
Kimmo

Reputation: 1966

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

Upvotes: 2

Xuan
Xuan

Reputation: 5629

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

Upvotes: 6

theller
theller

Reputation: 2879

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

Upvotes: 5

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