For developers

مايفيد الدومين لان يطلب رقم 🥲

import random,string,requests
from requests.packages.urllib3.exceptions import InsecureRequestWarning
requests.packages.urllib3.disable_warnings(InsecureRequestWarning)
Email="You Email"
SoS=0
while True:
    Port =random.randint(1024, 65535)
    Ip = f"{random.randint(1, 255)}.{random.randint(0, 255)}.{random.randint(0, 255)}.{random.randint(0, 255)}:"
    Proxy=str(Ip)+str(Port)
    letters_and_digits = string.ascii_lowercase + string.digits
    md5 = ''.join(random.choice(letters_and_digits) for _ in range(32))
    url = "https://alt-auth.mail.ru/api/v1/pushauth/info"
    payload = f"login={Email}%40mail.ru&md5_post_signature={md5}"
    headers = {
  'User-Agent': "okhttp/4.11.0",
  'Accept-Encoding': "gzip",
  'Content-Type': "application/x-www-form-urlencoded"
}
    response = requests.post(url, data=payload, verify=False,proxies={'http': Proxy}).text
    SoS+=1
    if '"available":true' in response:
        print(f'[{SoS}]Taken : {Email}')
    elif '"available":false' in response:
        print(f'[{SoS}]Available : {Email}')
    else:
        print(response)

كود فحص الايميل اذا مستخدم او لا على (Mail .Ru)

The Code Check Email Taken Or No In (Mail. Ru)

org = "topython@gmail.com"
res = "t*****n@gmail.com"

if (org.split("@")[0][0] == res.split("@")[0][0] and org.split("@")[0][-1] == res.split("@")[0][-1]) and \
   (org.split("@")[1] == res.split("@")[1] or (org.split("@")[1].split(".")[0][0] == res.split("@")[1].split(".")[0][0] and org.split("@")[1].split(".")[0][-1] == res.split("@")[1].split(".")[0][-1])):
    print(1)
else:
    print(0)

يفيدكم شرط اذا كان ريست ما يشبه الايميل ما يرسله

ضلعي انت تحتاج متابعين مو هاني صديقك 😪

اصعدو نحل مشاكل

⋘─────━𓆩𝐋7𝐍𓆪‏━─────⋙ Name : Username : astonbodywrap Email : astonbodywrap@gmail.com Followers : 0 Following : 0 Date : 2012 id : 191004296 Posts : 0 Bio : ASTONbodywrap/// Rests a*******p@gmail.com BY : @g_4_q ⋘─────━𓆩𝐋7𝐍𓆪‏━─────⋙

اريد شريف يفعلي مميز حتلو ب $ انزللكم ادوات ولا حد يدزلي اهداء @g_4_q

صديقي هاني يحتاج دعم قناته ما فيها متابعين 🙁 https://t.me/K_Q_A

Name : Username : lponghoy Email : lponghoy@gmail.com Followers : 0 Following : 0 Date : 2013 id : 363613954 Posts : 0 Bio : Rests l******y@gmail.com BY : @g_4_q

Name : Username : sherieeric40 Email : sherieeric40@gmail.com Followers : 0 Following : 0 Date : 2012 id : 243501594 Posts : 0 Bio : Rests s*******0@gmail.com BY : @g_4_q

Name : Username : stroonstr Email : stroonstr@gmail.com Followers : 0 Following : 0 Date : 2013 id : 303084926 Posts : 0 Bio : Rests s*******r@gmail.com BY : @g_4_q

امسحهم لحن هذن

ولك غير تشرح الهم

Fetch specific Instagram account information

from Topython import Instagram

info_ig = Instagram.information("Your Username")

print(info_ig) # Can You Use json