The text you are looking for is in the 2nd TR inside the table. So, print tbl.findAll(‘tr’)[2] There really isn’t any good structure inside that text, however, so you’re on your own about getting the contact name, etc. solved Why…
total = [] for line in lines: link = line.split(“ “)[1:] total.append(link) That should work. Let me know if that’s not what you were looking for. 1 solved storing each robots.txt value in an array
Use a dictionary, not variables. If you have variable names, the way to get the variable contents is evil, and almost always there is a better way. In this case: server_connection_commands = { “AdminServer”: “putty.exe -ssh 1.1.1.1”, “Server1”: “putty.exe -ssh…
The problem is, sizes doesn’t return anything. print does not return values, it just prints. You’d need to do something like: acc = “” for span in Size: acc += span.text.replace(‘EU:’,”).strip() # Concatenate the strings together return acc There are…
This is not the most straight forward post request, if you look in developer tools or firebug you can see the formdata from a successful browser post: All that is pretty straight forward bar the fact you see some :…
CONCLUSION – 07-25-2021 After looking at this problem in more detail, I believe that it is NOT technically possible to use Python Requests to scrape the website and table in your question. Which means that your question cannot be solved…
If you are getting error messages from the API like this: {“message”: “”, “code”: “api.404”, “data”: {}, “success”: false, “error”: false} try requesting each UUID seperately and once you have gotten a good response, try running it with your program.…
I figured it out: def add_photo(entry_id, image_path): return requests.post( url = URL, headers = HEADER, files = { ‘entry[entry_id]’: (None, entry_id, ‘text’), ‘entry[photo]’: (os.path.basename(image_path), open(image_path, ‘rb’), ‘image/jpg’, {‘Expires’: ‘0’}) } ) solved How to use python-request to post a file…
maybe this will help import requests city = input(“What is the name of the city?: “) url= “http://api.openweathermap.org/data/2.5/weather?q={city}&appid=*************”.format(city=city) response = requests.get(url) print(response) 2 solved Bad Request in Python
The element you’re interested in is dynamically generated, after the initial page load, which means that your browser executed JavaScript, made other network requests, etc. in order to build the page. Requests is just an HTTP library, and as such…
So off the top of my head you can look at 2 things. 1) Asyncio (be careful this example uses threading and is not thread safe specifically the function asyncio.gather) import asyncio for work in [1,2,3,4,5]: tasks.append(method_to_be_called(work)) results = await…
Inside your for loop you’re creating a new Session object – you should only have one (you have one at the start of your code) You’re also using a .get() request when it should be a .post() replace: # Getting…
Solved mydata = {’email’: form.vars.email, ‘password’:form.vars.password} resp= requests.post(‘url’, json = mydata) I don’t know why but I had already tried it and it didn’t work solved Python and curl equivalent
Try this: import time import requests link = ” params = {‘user’: ‘0xe744d23107c9c98df5311ff8c1c8637ec3ecf9f3’} payload = {“size”: 20} with requests.Session() as s: s.headers[‘User-Agent’] = ‘Mozilla (Windows NT 6.1) AppleWebKit (KHTML, like Gecko) Chrome Safari’ res = s.post(link,params=params,json=payload) for item in res.json():…