CODEWITHSUNDEEP
pythonparametersurlliburllib2vapix
Dillon Tucker
Dillon Tucker

Reputation: 55

Python - urllib - Server is not getting everything after '?' in URL

I am writing a Python3 script to trigger an Axis camera via their HTTP API. The string they need in the url to trigger the Digital Input (with authorization) is:

"http://{ip}/axis-cgi/param.cgi?action=update&IOPort.I0.Input.Trig={open/closed}"

If I put this in a browser it works AKA response 200 - OK. When I run this on my Linux machine using the urllib request.

#encode user,pass
values = { 'username': username,'password': password }
data = urllib.urlencode(values)

#Axis param.cgi action
action = "update"
trig = "closed"

cgi_args = {"action": action, "IOPort.I0.Input.Trig": trig}
cgi_data = urllib.urlencode(cgi_args)
url = "http://{ip}/axis-cgi/param.cgi?{data}".format(ip=ip, data=cgi_data)

req = urllib2.Request(url, data, headers={'Content-type': 'text/html'})
print(req.get_full_url())

response = urllib2.urlopen(req)
result = response.read()
print (result)

The output is:

http://192.168.50.191/axis-cgi/param.cgi?action=update&IOPort.I0.Input.Trig=closed
action must be specified

I know that I am authenticated otherwise I get Unauthorized response from server.

As you can see for debugging I print the req.get_full_url() to make sure I've built the url string correctly, however the server responds with action must be specified which is what I get in the browser when I just have http://192.168.50.191/axis-cgi/param.cgi? in the address bar. So the action portion of the URL appears to not be making it to the server.

I've tried:

  1. Using %3F as the ? character, but I get 404 error
  2. Embedding the data in the data parameter does not work either

Anything I am missing here?

Upvotes: 0

Views: 391

Answers (1)

Dillon Tucker
Dillon Tucker

Reputation: 55

used pycurl with digest authorization for a GET request with auth:

def configCurl():
username = "user"
password = "pass"
auth_mode = pycurl.HTTPAUTH_DIGEST
curl = pycurl.Curl()
curl.setopt(pycurl.HTTPAUTH, auth_mode)
curl.setopt(pycurl.USERPWD, "{}:{}".format(username, password))
curl.setopt(pycurl.WRITEFUNCTION, lambda x: None)
return curl

curl = configCurl()
curl.setopt(curl.URL, url)
curl.perform()
curl.close()

Upvotes: 0

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