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  1. Hello, I'm trying to make a league of legends checker/cracker in python. I have all ready and running, the problem is that its too slow. Here are my steps: - Do a GET to to get the needed cookies. - Do a put to with the username and password. For some reason this takes too long and my program is doing like 3 CPM even with multiple threads at the same time. Has anyone a faster way to check if the username and password are correct? And what other recommendations can you give me to make it faster? Thanks and sorry for bad English.
  2. Lets say you want to crack accounts for a page called What you have to do is: - Find the address where username and password are being send to check if they are correct or not: You can do this doing inspect element on the page, then click on network section, there you will see all the traffic that is being send from that page. Then click login button and find out where they are being send. You can learn more on how to do this on any youtube tutorial. - Make the python application. You will only need requests library for a simple cracker: You have to send the user and password to the URL that you find in the previous step using the correct method (post or get) Then you probably going to get an answer from the server saying if the password was correct. Some pages do this with a 200 code, others use different methods, you have to find witch is being used on your website. And that is basically it. If you didn't understand what i write you can check this tutorial that i found on internet: Here is an example: import Requests payload = {"username":"yourusername","password":"yourpassword"} login ="", json=payload) if login.status_code == "200": print("Correct username and password") else: print("Incorrect username and password") In some websites you will need to change your user agent and keep some cookies because they will know you are not a real user and block you (you can find out how to do this on internet, it's not hard). Good luck and sorry for my bad English.
  3. thanks bro i love you please