I am fairly a noob at this and have been trying to use requests modules to post a multipart/form-data. To clarify, the exact test case I am trying to use is the one same as in https://github.com/kennethreitz/requests/issues/1081. i.e. I am trying to do a post without a file :
--3eeaadbfda0441b8be821bbed2962e4d
Content-Disposition: form-data; name="key1"
value1
--3eeaadbfda0441b8be821bbed2962e4d
As per the discussion on the thread, I tried MultiPart form data scheme to do the following:
import requests
from requests_data_schemes import multipart_formdata as mfd
post_data = [('mouseAction', 'toggle'), ('zone' ,'10')]
post_data = mfd(post_data)
headers = {'Content-Type': 'multipart/form-data'}
req = requests.post(<url>, data=post_data, headers=headers)
However, the test server is throwing me an error saying that it cannot detect the boundary of the multipart form data.
I tried providing the boundary in the header too, but apparently its not working.
boundary = post_data[2: post_data.find('\r\n')]
headers = {'Content-Type': 'multipart/form-data; boundary={}'.format(boundary)}
Am i missing something simple?
P.S. From a bit of surfing I found a few solutions using base urllib2 but that would be my last resort as requests lets me do a lot things pretty easily.
Yeah this is a bug I have to address. You're better off at this point doing the following:
from requests.packages.urllib3.filepost import encode_multipart_formdata
(content, header) = encode_multipart_formdata([('key', 'value')])
r = requests.post(url, data=content, headers={'Content-Type': header})
Related
I try to read JSON-formatted data from the following public URL: http://ws-old.parlament.ch/factions?format=json. Unfortunately, I was not able to convert the response to JSON as I always get the HTML-formatted content back from my request. Somehow the request seems to completely ignore the parameters for JSON formatting passed with the URL:
import urllib.request
response = urllib.request.urlopen('http://ws-old.parlament.ch/factions?format=json')
response_text = response.read()
print(response_text) #why is this HTML?
Does somebody know how I am able to get the JSON formatted content as displayed in the web browser?
You need to add "Accept": "text/json" to request header.
For example using requests package:
r = requests.get(r'http://ws-old.parlament.ch/factions?format=json',
headers={'Accept':'text/json'})
print(r.json())
Result:
[{'id': 3, 'updated': '2022-02-22T14:59:17Z', 'abbreviation': ...
Sorry for you but these web services have a misleading implementation. The format query parameter is useless. As pointed out by #maciek97x only the header Accept: <format> will be considered for the formatting.
So your can directly call the endpoint without the ?format=json but with the header Accept: text/json.
I am trying to write a Python script that will allow me to accomplish what I normally would by using CURL to perform an API "GET". I have browsed through some questions on here but I am still a bit confused on how to do this; from what I have seen, I can use the "requests" library like this:
URL = requests.get("www.example.com/123")
However, the Curl command I normally run uses authentication as well. Here is an example:
curl -X GET -H 'Authorization: hibnn:11111:77788777YT666:CAL1' 'http://api.example.com/v1.11/user?id=123456
Can I still use the requests library when creating a script to pull this data? Specifically, I would need to be able to pass along the authentication info along with the URL itself.
Update:
Here is what my code looks like:
import requests
import json
url = ("api.example.com/v1.11/user?id=123456")
header = {"Authorization": "hibnn:11111:77788777YT666:CAL1"}
response = requests.get(url, headers=header)
print(response.json)
However, I am getting a response [401] error. I think I am probably doing something wrong with my headers, can anyone help?
You may pass headers as keyword argument to the get method.
>>> url = 'https://api.github.com/some/endpoint'
>>> headers = {'user-agent': 'my-app/0.0.1'}
>>> r = requests.get(url, headers=headers)
Reference
Today I actually needed to retrieve data from the http-header response. But since I've never done it before and also there is not much you can find on Google about this. I decided to ask my question here.
So actual question: How does one print the http-header response data in python? I'm working in Python3.5 with the requests module and have yet to find a way to do this.
Update: Based on comment of OP, that only the response headers are needed. Even more easy as written in below documentation of Requests module:
We can view the server's response headers using a Python dictionary:
>>> r.headers
{
'content-encoding': 'gzip',
'transfer-encoding': 'chunked',
'connection': 'close',
'server': 'nginx/1.0.4',
'x-runtime': '148ms',
'etag': '"e1ca502697e5c9317743dc078f67693f"',
'content-type': 'application/json'
}
And especially the documentation notes:
The dictionary is special, though: it's made just for HTTP headers. According to RFC 7230, HTTP Header names are case-insensitive.
So, we can access the headers using any capitalization we want:
and goes on to explain even more cleverness concerning RFC compliance.
The Requests documentation states:
Using Response.iter_content will handle a lot of what you would otherwise have to handle when using Response.raw directly. When streaming a download, the above is the preferred and recommended way to retrieve the content.
It offers as example:
>>> r = requests.get('https://api.github.com/events', stream=True)
>>> r.raw
<requests.packages.urllib3.response.HTTPResponse object at 0x101194810>
>>> r.raw.read(10)
'\x1f\x8b\x08\x00\x00\x00\x00\x00\x00\x03'
But also offers advice on how to do it in practice by redirecting to a file etc. and using a different method:
Using Response.iter_content will handle a lot of what you would otherwise have to handle when using Response.raw directly
How about something like this:
import urllib2
req = urllib2.Request('http://www.google.com/')
res = urllib2.urlopen(req)
print res.info()
res.close();
If you are looking for something specific in the header:
For Date: print res.info().get('Date')
Here's how you get just the response headers using the requests library like you mentioned (implementation in Python3):
import requests
url = "https://www.google.com"
response = requests.head(url)
print(response.headers) # prints the entire header as a dictionary
print(response.headers["Content-Length"]) # prints a specific section of the dictionary
It's important to use .head() instead of .get() otherwise you will retrieve the whole file/page like the rest of the answers mentioned.
If you wish to retrieve a URL that requires authentication you can replace the above response with this:
response = requests.head(url, auth=requests.auth.HTTPBasicAuth(username, password))
easy
import requests
site = "https://www.google.com"
headers = requests.get(site).headers
print(headers)
if you want something specific
print(headers["domain"])
I'm using the urllib module, with the following code:
from urllib import request
with request.urlopen(url, data) as f:
print(f.getcode()) # http response code
print(f.info()) # all header info
resp_body = f.read().decode('utf-8') # response body
its very easy u can type
print(response.headers)
or my fav
print(requests.get('url').headers)
also u can use
print(requests.get('url').content)
Try to use req.headers and that's all. You will get the response headers ;)
import pprint
import requests
res = requests.request("GET", "https://google.com")
pprint.PrettyPrinter(indent=2).pprint(dict(res.headers))
rocksteady's solution worked
He did originally refer to dictionaries. But the following code to send the JSON string also worked wonders using requests:
import requests
headers = {
'Authorization': app_token
}
url = api_url + "/b2api/v1/b2_get_upload_url"
content = json.dumps({'bucketId': bucket_id})
r = requests.post(url, data = content, headers = headers)
I'm working with an API that requires me to send JSON as a POST request to get results. Problem is that Python 3 won't allow me to do this.
The following Python 2 code works fine, in fact it's the official sample:
request = urllib2.Request(
api_url +'/b2api/v1/b2_get_upload_url',
json.dumps({ 'bucketId' : bucket_id }),
headers = { 'Authorization': account_authorization_token }
)
response = urllib2.urlopen(request)
However, using this code in Python 3 only makes it complain about data being invalid:
import json
from urllib.request import Request, urlopen
from urllib.parse import urlencode
# -! Irrelevant code has been cut out !-
headers = {
'Authorization': app_token
}
url = api_url + "/b2api/v1/b2_get_upload_url"
# Tested both with encode and without
content = json.dumps({'bucketId': bucket_id}).encode('utf-8')
request = Request(
url=url,
data=content,
headers=headers
)
response = urlopen(req)
I've tried doing urlencode(), like you're supposed to. But this returns a 400 status code from the web server, because it's expecting pure JSON. Even if the pure JSON data is invalid, I need to somehow force Python into sending it.
EDIT: As requested, here are the errors I get. Since this is a flask application, here's a screenshot of the debugger:
Screenshot
Adding .encode('utf-8') gives me an "Expected string or buffer" error
EDIT 2: Screenshot of the debugger with .encode('utf-8') added
Since I have a similar application running, but the client still was missing, I tried it myself.
The server which is running is from the following exercise:
Miguel Grinberg - designing a restful API using Flask
That's why it uses authentication.
But the interesting part: Using requests you can leave the dictionary as it is.
Look at this:
username = 'miguel'
password = 'python'
import requests
content = {"title":"Read a book"}
request = requests.get("http://127.0.0.1:5000/api/v1.0/projects", auth=(username, password), params=content)
print request.text
It seems to work :)
Update 1:
POST requests are done using requests.post(...)
This here describes it well : python requests
Update 2:
In order to complete the answer:
requests.post("http://127.0.0.1:5000/api/v1.0/projects", json=content)
sends the json-string.
json is a valid parameter of the request and internally uses json.dumps()...
I have another question about posts.
This post should be almost identical to one referenced on stack overflow using this question 'Using request.post to post multipart form data via python not working', but for some reason I can't get it to work. The website is http://www.camp.bicnirrh.res.in/predict/. I want to post a file that is already in the FASTA format to this website and select the 'SVM' option using requests in python. This is based on what #NorthCat gave me previously, which worked like a charm:
import requests
import urllib
file={'file':(open('Bishop/newdenovo2.txt','r').read())}
url = 'http://www.camp.bicnirrh.res.in/predict/hii.php'
payload = {"algo[]":"svm"}
raw = urllib.urlencode(payload)
response = session.post(url, files=file, data=payload)
print(response.text)
Since it's not working, I assumed the payload was the problem. I've been playing with the payload, but I can't get any of these to work.
payload = {'S1':str(data), 'filename':'', 'algo[]':'svm'} # where I tried just reading the file in, called 'data'
payload = {'svm':'svm'} # not actually in the headers, but I tried this too)
payload = {'S1': '', 'algo[]':'svm', 'B1': 'Submit'}
None of these payloads resulted in data.
Any help is appreciated. Thanks so much!
You need to set the file post variable name to "userfile", i.e.
file={'userfile':(open('Bishop/newdenovo2.txt','r').read())}
Note that the read() is unnecessary, but it doesn't prevent the file upload succeeding. Here is some code that should work for you:
import requests
session = requests.session()
response = session.post('http://www.camp.bicnirrh.res.in/predict/hii.php',
files={'userfile': ('fasta.txt', open('fasta.txt'), 'text/plain')},
data={'algo[]':'svm'})
response.text contains the HTML results, save it to a file and view it in your browser, or parse it with something like Beautiful Soup and extract the results.
In the request I've specified a mime type of "text/plain" for the file. This is not necessary, but it serves as documentation and might help the receiving server.
The content of my fasta.txt file is:
>24.6jsd2.Tut
GGTGTTGATCATGGCTCAGGACAAACGCTGGCGGCGTGCTTAATACATGCAAGTCGAACGGGCTACCTTCGGGTAGCTAGTGGCGGACGGGTGAGTAACACGTAGGTTTTCTGCCCAATAGTGGGGAATAACAGCTCGAAAGAGTTGCTAATACCGCATAAGCTCTCTTGCGTGGGCAGGAGAGGAAACCCCAGGAGCAATTCTGGGGGCTATAGGAGGAGCCTGCGGCGGATTAGCTAGATGGTGGGGTAAAGGCCTACCATGGCGACGATCCGTAGCTGGTCTGAGAGGACGGCCAGCCACACTGGGACTGAGACACGGCCCAGACTCCTACGGGAGGCAGCAGTAAGGAATATTCCACAATGGCCGAAAGCGTGATGGAGCGAAACCGCGTGCGGGAGGAAGCCTTTCGGGGTGTAAACCGCTTTTAGGGGAGATGAAACGCCACCGTAAGGTGGCTAAGACAGTACCCCCTGAATAAGCATCGGCTAACTACGTGCCAGCAGCCGCGGTAATACGTAGGATGCAAGCGTTGTCCGGATTTACTGGGCGTAAAGCGCGCGCAGGCGGCAGGTTAAGTAAGGTGTGAAATCTCCCTGCTCAACGGGGAGGGTGCACTCCAGACTGACCAGCTAGAGGACGGTAGAGGGTGGTGGAATTGCTGGTGTAGCGGTGAAATGCGTAGAGATCAGCAGGAACACCCGTGGCGAAGGCGGCCACCTGGGCCGTACCTGACGCTGAGGCGCGAAGGCTAGGGGAGCGAACGGGATTAGATACCCCGGTAGTCCTAGCAGTAAACGATGTCCACTAGGTGTGGGGGGTTGTTGACCCCTTCCGTGCCGAAGCCAACGCATTAAGTGGACCGCCTGGGGAGTACGGTCGCAAGACTAAAACTCAAAGGAATTGACGGGGACCCGCACAAGCAGCGGAGCGTGTGGTTTAATTCGATGCGACGCGAAGAACCTTACCTGGGCTTGACATGCTATCGCAACACCCTGAAAGGGGTGCCTCCTTCGGGACGGTAGCACAGATGCTGCATGGCTGTCGTCAGCTCGTGTCGTGAGATGTTGGGTTAAGTCCCGCAACGAGCGCAACCCCTGTCCTTAGTTGTATATCTAAGGAGACTGCCGGAGACAAACCGGAGGAAGGTGGGGATGACGTCAAGTCAGCATGGCTCTTACGTCCAGGGCTACACATACGCTACAATGGCCGTTACAGTGAGATGCCACACCGCGAGGTGGAGCAGATCTCCAAAGGCGGCCTCAGTTCAGATTGCACTCTGCAACCCGAGTGCATGAAGTCGGAGTTGCTAGTAACCGCGTGTCAGCATAGCGCGGTGAATATGTTCCCGGGTCTTGTACACACCGCCCGTCACGTCATGGGAGCCGGCAACACTTCGAGTCCGTGAGCTAACCCCCCCTTTCGAGGGTGTGGGAGGCAGCGGCCGAGGGTGGGGCTGGTGACTGGGACGAAGTCGTAACAAGGT