Filecrypt Bypass - get direct links of filecrypt links

filecrypt.py

from bs4 import BeautifulSoup
import cloudscraper
import requests
import json

url = input("enter filecrypt link like https://filecrypt.co/Container/73F6D9D43B.html or html filepath like 75C3806BBE.html: ")
        
def getlinks(dlc):
    headers = {
    'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64; rv:103.0) Gecko/20100101 Firefox/103.0',
    'Accept': 'application/json, text/javascript, */*',
    'Accept-Language': 'en-US,en;q=0.5',
    'X-Requested-With': 'XMLHttpRequest',
    'Origin': 'http://dcrypt.it',
    'Connection': 'keep-alive',
    'Referer': 'http://dcrypt.it/',
    }

    data = {
        'content': dlc,
    }

    response = requests.post('http://dcrypt.it/decrypt/paste', headers=headers, data=data).json()["success"]["links"]
    links = ""
    for link in response:
        links = links + link + "\n\n"
    print(links)
    return links

def filecrypt(url):

    client = cloudscraper.create_scraper(allow_brotli=False)
    headers = {
    "authority": "filecrypt.co",
    "accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9",
    "accept-language": "en-US,en;q=0.9",
    "cache-control": "max-age=0",
    "content-type": "application/x-www-form-urlencoded",
    "origin": "https://filecrypt.co",
    "referer": url,
    "user-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/105.0.0.0 Safari/537.36" 
    }
    
    resp = client.get(url, headers=headers)
    soup = BeautifulSoup(resp.content, "html.parser")

    buttons = soup.find_all("button")
    for ele in buttons:
        line = ele.get("onclick")
        if line !=None and "DownloadDLC" in line:
            dlclink = "https://filecrypt.co/DLC/" + line.split("DownloadDLC('")[1].split("'")[0] + ".html"
            break

    resp = client.get(dlclink,headers=headers)
    if resp.text == "":
        print("Falied, if this is a link then try downloading html file of that link and pass the filepath, if it is a filepath then redownload html file and retry")
    else:
        getlinks(resp.text)       
       
filecrypt(url)

@jeovaemanuel it's need cookies at header, you can add cookie value on filecrypt(url) def.
e.g :
"cookie": "PHPSESSID=dgad97gqjjpgiae83qjdddv3a2; lang=en; ab=1696546708; ha=1; %3A_$Yo0FsXyVd1CoS01LMTUFWfviO5ANC18Z.c=8; haad=1; haac=1; ac=22; acn=22",

cookies expire, we cant hardcode it