使用 Python 理解 API 并与之交互

Question

I have a basic API installed as my localhost server that does functions such as add camera, star camera, list cameras, snapshot of camera frame, etc.

My problem is after following the documentation I still can't seem to interact with it well and get the response I need. Here is the code I use to log in and get validation token:

import requests
import urllib.request
import json


base_url = "http://localhostip:8080/api/user/login?"
parameters = {
    "username": username,
    "password": password
}
auth_tok = requests.post(base_url + urllib.parse.urlencode(parameters)).json()
print(auth_tok)

I get the correct documented response with a token, so following the documentation to add camera I need 2 parameters, URL and Name , so I did:

base_url = "http://localhostip:8080/api/camera/add?"

parameters = {

    "url": 'rtsp://192.168.1.23/1', 

    #or video file 
    "url" : '/home/video/sample.mov'

    "name" : 'cam1'

}
r = requests.post(base_url + urllib.parse.urlencode(parameters),headers={'Authorization': auth_tok})

when I print the response:

-print (r)
-print (r.url)
-print(r.status_code)
-print(r.json())

I get this:

<Response [500]>

http://192.168.0.162:8080/service/api/camera/add?url=rtsp%3A%2F%2Frtsp%3A%2F%2F192.168.1.23&name=cam1

500

{'code': -111, 'message': None}

According to documentation the correct url should be like this:

http://192.168.0.6:8080/service/api/camera/add?url=rtsp://192.168.1.23&name=cam1

and the response should be:

Response: {"status":"ok"}

So why and how to make the URL POST in the correct format, because I suspect this is the issue, the URL has these encoding symbols that may be messing up the request?

When I use the web browser GUI of this API I can add the camera or even a video file to play but I'm trying to do the same with Python so I can do further processing in future.

Answer 1

Your problem is when you encode the ' / / ' symbol, so, in order to fix that, you need to use another function from urllib, urllib.parse.unquote(), and use as parameter your encoding function urllib.parse.urlencode(parameters):

import urllib

parameters = {
    "url": 'rtsp://192.168.1.23/1',
    "name" : 'cam1'
}

The results are :

print(urllib.parse.urlencode(parameters))
'url=rtsp%3A%2F%2F192.168.1.23%2F1&name=cam1'

print(urllib.parse.unquote(urllib.parse.urlencode(parameters)))
'url=rtsp://192.168.1.23/1&name=cam1'

Source https://docs.python.org/3.0/library/urllib.parse.html#urllib.parse.unquote