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[Python] Selenium WebDriver Selector Troubles

I'm trying to fucking sign up to "og users" (do a google search to find the forum, not allowed to link) and when trying to add my main account as the referral page, I can't even get on that text box. (look at the sign up page to understand lol).

Could anybody hook me up with the source select it and enter text, thanks lol.

PS:: I tried a bunch of shit trust me...

Will rep whoever helps/attempts to help of course. 

Anyways, thanks HF. I love u all   ✌

[Image: 9Y60Xjq.png]
htt ps:/ /pastebin.com/BTmU9hSW

This should work.


Of course this is a just a sample. You should edit the header, add Proxies etc.
But well.. it's not that hard.

Good luck!

btw. the bot protection is hilarious :D

2nd btw. don't use "time.sleep(x)"! Use "finally" instead, because you don't know how long it take to load the site. I used time.sleep, because it's faster to type. Just as I said.. just an example.
Hey! I'm quite sorry for commenting on here but because of my negative rep i couldn't on your other thread.
I am highly interested in getting mentored and have allot of time to spare!
And i am devoted to something i start with.
(Dec 1, 2017 01:59 PM)TheKid_ Wrote:  Hey! I'm quite sorry for commenting on here but because of my negative rep i couldn't on your other thread.
I am highly interested in getting mentored and have allot of time to spare!
And i am devoted to something i start with.

Who do you mean?
no idea if this will work, selenium isn't working for me anymore
Spoiler (Click to View)
i like python lua js
I liked scrubjay's answer, so I made a shorter one.
https://hastebin.com/qixadeyuso.py
It can bypass the COPPA registration, and anti-bot.
I've done it something like this, to make it more random. Uses a random username, password and email every registration. You could use the Fake-Agent library to get some fake user-agents everytime you run it. Besides that you could load a proxy list in it to use a random proxy every time.

Here is my pastebin URL; https://pastebin.com/Uyf1s2qC


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