Implementation HMAC-SHA1 in python

PythonOauthSha1Hmac

Python Problem Overview


I am trying to use the OAuth of a website, which requires the signature method to be 'HMAC-SHA1' only.

I am wondering how to implement this in Python?

Python Solutions


Solution 1 - Python

Pseudocodish:

def sign_request():
    from hashlib import sha1
    import hmac

    # key = b"CONSUMER_SECRET&" #If you dont have a token yet
    key = b"CONSUMER_SECRET&TOKEN_SECRET" 


    # The Base String as specified here: 
    raw = b"BASE_STRING" # as specified by OAuth
       
    hashed = hmac.new(key, raw, sha1)
    
    # The signature
    return hashed.digest().encode("base64").rstrip('\n')

Signature errors usually reside in the base-string, make sure you understand this (as stated by the OAuth1.0 spec here: https://datatracker.ietf.org/doc/html/draft-hammer-oauth-10#section-3.4.1).

The following inputs are used to generate the Signature Base String:

  1. HTTP Method (for example GET)

  2. Path (for example http://photos.example.net/photos)

  3. Parameters, alphabetically, such as (line breaks for readability):

     file=vacation.jpg
     &oauth_consumer_key=dpf43f3p2l4k3l03
     &oauth_nonce=kllo9940pd9333jh
     &oauth_signature_method=HMAC-SHA1
     &oauth_timestamp=1191242096
     &oauth_token=nnch734d00sl2jdk
     &oauth_version=1.0
     &size=original
    

Concatenate and URL encode each part and it ends up as:

GET&http%3A%2F%2Fphotos.example.net%2Fphotos&file%3Dvacation.jpg%26 oauth_consumer_key%3Ddpf43f3p2l4k3l03%26oauth_nonce%3Dkllo9940pd9333jh%26 oauth_signature_method%3DHMAC-SHA1%26oauth_timestamp%3D1191242096%26 oauth_token%3Dnnch734d00sl2jdk%26oauth_version%3D1.0%26size%3Doriginal

Solution 2 - Python

For the love of God, if you do ANYTHING with oauth, use the requests library for Python! I tried to implement HMAC-SHA1 using the hmac library in Python and it's a lot of headaches, trying to create the correct oauth base string and such. Just use requests and it's as simple as:

>>> import requests
>>> from requests_oauthlib import OAuth1

>>> url = 'https://api.twitter.com/1.1/account/verify_credentials.json'
>>> auth = OAuth1('YOUR_APP_KEY', 'YOUR_APP_SECRET', 'USER_OAUTH_TOKEN', 'USER_OAUTH_TOKEN_SECRET')

>>> requests.get(url, auth=auth)

Requests Authentication

[Requests Oauth Library] 2

Solution 3 - Python

You can try following method.

def _hmac_sha1(input_str):
        raw = input_str.encode("utf-8")
        key = 'your_key'.encode('utf-8')
        hashed = hmac.new(key, raw, hashlib.sha1)
        return base64.encodebytes(hashed.digest()).decode('utf-8')

Solution 4 - Python

Solution 5 - Python

Finally here's an actually working solution (tested with Python 3) utilizing oauthlib.

I use the first OAuth step given as an example in the official RTF 1:

Client Identifier: dpf43f3p2l4k3l03
Client Shared-Secret: kd94hf93k423kf44

POST /initiate HTTP/1.1
Host: photos.example.net
Authorization: OAuth realm="Photos",
    oauth_consumer_key="dpf43f3p2l4k3l03",
    oauth_signature_method="HMAC-SHA1",
    oauth_timestamp="137131200",
    oauth_nonce="wIjqoS",
    oauth_callback="http%3A%2F%2Fprinter.example.com%2Fready",
    oauth_signature="74KNZJeDHnMBp0EMJ9ZHt%2FXKycU%3D"

The value for oauth_signature is what we would like to calculate.

The following defines what we want to sign:

# There is no query string present.
# In case of http://example.org/api?a=1&b=2 - the value
# would be "a=1&b=2".
uri_query=""

# The oauthlib function 'collect_parameters' automatically
# ignores irrelevant header items like 'Content-Type' or
# 'oauth_signature' in the 'Authorization' section.
headers={
    "Authorization": (
        'OAuth realm="Photos", '
        'oauth_nonce="wIjqoS", '
        'oauth_timestamp="137131200", '
        'oauth_consumer_key="dpf43f3p2l4k3l03", '
        'oauth_signature_method="HMAC-SHA1", '
        'oauth_callback="http://printer.example.com/ready"'
    )
}

# There's no POST data here - in case it was: x=1 and y=2,
# then the value would be '[("x","1"),("y","2")]'.
data=[]

# This is the above specified client secret which we need
# for calculating the signature.
client_secret="kd94hf93k423kf44"

And here we go:

import oauthlib.oauth1.rfc5849.signature as oauth

params = oauth.collect_parameters(
    uri_query="",
    body=data, 
    headers=headers,
    exclude_oauth_signature=True, 
    with_realm=False
)

norm_params = oauth.normalize_parameters(params)

base_string = oauth.construct_base_string(
    "POST", 
    "https://photos.example.net/initiate", 
    norm_params
)

sig = oauth.sign_hmac_sha1(
    base_string, 
    client_secret, 
    '' # resource_owner_secret - not used
)

from urllib.parse import quote_plus

print(sig)
# 74KNZJeDHnMBp0EMJ9ZHt/XKycU=

print(quote_plus(sig))
# 74KNZJeDHnMBp0EMJ9ZHt%2FXKycU%3D

Solution 6 - Python

There are multiple python libraries available at the oauth website, but if you're just interested in a specific implementation you could have a look at one of them.

Solution 7 - Python

In Python 3.7 there is an optimized way to do this. HMAC(key, msg, digest).digest() uses an optimized C or inline implementation, which is faster for messages that fit into memory.

> Return digest of msg for given secret key and digest. The function is > equivalent to HMAC(key, msg, digest).digest(), but uses an optimized C > or inline implementation, which is faster for messages that fit into > memory. The parameters key, msg, and digest have the same meaning as > in new(). > > CPython implementation detail, the optimized C implementation is only > used when digest is a string and name of a digest algorithm, which is > supported by OpenSSL.

https://docs.python.org/3/library/hmac.html#hmac.digest

Solution 8 - Python

In Python 3.7 there is an optimized way to do this. HMAC(key, msg, digest).digest() uses an optimized C or inline implementation, which is faster for messages that fit into memory.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionxiaohan2012View Question on Stackoverflow
Solution 1 - PythonJon NylanderView Answer on Stackoverflow
Solution 2 - PythonBlairg23View Answer on Stackoverflow
Solution 3 - PythonQy ZuoView Answer on Stackoverflow
Solution 4 - PythonAndrey AtapinView Answer on Stackoverflow
Solution 5 - PythonRaffaelView Answer on Stackoverflow
Solution 6 - PythonTillView Answer on Stackoverflow
Solution 7 - PythonSuperNovaView Answer on Stackoverflow
Solution 8 - Pythonbồn vệ sinhView Answer on Stackoverflow