CODEWITHSUNDEEP
pythonhttphttp-headers
BajajG
BajajG

Reputation: 2174

Python requests.exception.ConnectionError: connection aborted "BadStatusLine"

I am trying to use the Python requests module to issue Http GET commands to access some REST based APIs. The urls are working fine on a RESTClient but when I use the same url in python, I get a connection error.

The code I am trying to execute is:

payload={"mac":new_mac,"token":token}
userloginurl="http://192.168.1.40:9119/uid"
r=requests.get(userloginurl,params=payload)
print(r.url)

If I test this url using RESTClient, I get a 200 OK status code in the response header along with some more fields. enter image description here But this is not working with python requests. The traceback of the error is shown below:

Traceback (most recent call last):
File "getAPids.py", line 34, in <module>
r=requests.get(userloginurl,params=payload)
  File "C:\Users\garvitab\python\lib\site-packages\requests\api.py", line 65, in
 get
return request('get', url, **kwargs)
File "C:\Users\garvitab\python\lib\site-packages\requests\api.py", line 49, in
request
response = session.request(method=method, url=url, **kwargs)
File "C:\Users\garvitab\python\lib\site-packages\requests\sessions.py", line 4
61, in request
resp = self.send(prep, **send_kwargs)
File "C:\Users\garvitab\python\lib\site-packages\requests\sessions.py", line 5
73, in send
r = adapter.send(request, **kwargs)
 File "C:\Users\garvitab\python\lib\site-packages\requests\adapters.py", line 4
15, in send
raise ConnectionError(err, request=request)
requests.exceptions.ConnectionError: ('Connection aborted.', BadStatusLine("''",
))

I looked up for the cause of the problem. Possibly, the response received is not formatted correctly. Is there a way to handle this problem?

Thanks in advance.

Upvotes: 39

Views: 116527

Answers (4)

Daniel Olson
Daniel Olson

Reputation: 531

In my case i was sending files of to large a size. I ended up streaming them and if they were still too large I decreased the size if there's and error:

this is sent to my own server and literally copy and pasted so you may not have _first or may have to change some things:

def upload_to_server(url, filedir, filename):
    print(f'uploading {k}')
    upload_file_path = os.path.join(filedir, filename)

    def read_in_chunks(file_object, chunk_size=524288):
        while True:
            data = file_object.read(chunk_size)
            if not data:
                break
            yield data

    trys = 5
    for i in range(trys):
        try:
            _first = 'true'
            with open(upload_file_path, 'rb') as f:
                for piece in read_in_chunks(f, chunk_size=int(524288 / (i + 1))):
                    r = requests.post(
                        url + '/upload/stream',
                        files={filename: piece)
                               },
                        headers={'first': _first})
                    # print('piece sent')
                    _first = 'false'
                    # print(r)
                    # print(r.text)
            break
        except ConnectionError as e:
            print(e)

    print(f'{k} was uploaded')

Upvotes: 0

cloud_IaaS
cloud_IaaS

Reputation: 101

I was getting the same error and spent hours on it. I found that you cannot call the flask server in a client using "Localhost". It has to be "127.0.0.1"

#server
from flask import Flask
app = Flask(__name__)

@app.route('/')
def hello_world():
   return 'Hello World'

if __name__ == '__main__':
   app.run(debug=True)

Now the client causing the error:

#client
import requests
x = requests.get('http://localhost:5000') # change it to 127.0.0.1
print(x.text)

Upvotes: 1

BajajG
BajajG

Reputation: 2174

The problem was with the url. This connection was meant to be established over https and I was using http in the python script. Hence the issue.

Upvotes: 42

Michael
Michael

Reputation: 7736

Have you actually checked, what gets send over the wire? I suppose you might have to convert your dictionary to a JSON string by yourself, or use the json= keyword instead of payload=. See http://docs.python-requests.org/en/latest/user/quickstart/#custom-headers for details. This may do the trick:

import json
payload = json.dumps({"mac":new_mac,"token":token})
userloginurl = "http://192.168.1.40:9119/uid"
r = requests.get(userloginurl, data=payload)
print(r.url)

Upvotes: 3

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